Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate the pressure exerted by one gram of helium present in a 5 L vessel at room temperature? |
|
Answer» Solution :The ideal GAS equation is given as, PV=nRT Number of MOLES `= N = 1/4 = 0.25` R = Universal gas constant `= 0.0821 L ATM K^(-1) mol^(-1)` V=Volume = 5 lit Pressure exerted by helium gas `P = (nRT)/(V) = (0.25 xx 0.0821 xx 298)/(5)` ` = 1.224 atm = 930 ` torr |
|
| 2. |
Calculate the pressure exerted by 5 moles of CO_2 in one litre vessel at 47^@C using Vanderwaal's equation. Also report the pressure of gas if it behaves ideally in nature. Given that a=3.592 atm lit^2 mol^(-2).b = 0.0427 lit mol^(-1) |
|
Answer» <P> SOLUTION :`P_("REAL") = 77.2` ATM`P_("IDEAL") = 131.36 atm |
|
| 3. |
Calculate the pressure exerted by 2 moles of sulphur hexafluoride in a steel vessel of volume 6 dm^(3) at 70^(@)C assuming it is an ideal gas. |
| Answer» Solution :We will USE the ideal gas equation for this CALCULATION as below:`P=(nRT)/(V)=(2molxx0.0821 "L atm "K^(-1).MMOL^(-1)xx(70+273K))/(6dm)=9.39 AYM` | |
| 4. |
Calculate the pressure exerted by 110 g of carbon dioxide in a vessel of 2 L capacity at 37^(@)C. Given that the van der Waals constants are a=3.59 L^(2) atm mol^(-2) and b=0.0427 L mol^(-1). Compare the value with the calculated value if the gas were considered as ideal. |
|
Answer» SOLUTION :According to VAN der Waals equation, `(p+(an^(2))/(V^(2)))(v-nb)=nRT"or"P=(nRT)/(V-nb)-(an^(2))/(V^(2))` Here, `N=(110)/(44)=2.5` moles. Substituting the given values,we get `P=((2.5 MOL)(0.0821" L atm "K^(-1) mol^(-1)))/((2 L-2.5 molxx0.0427 L mol^(-1)))-((3.59 L^(2) atm" mol"^(-2))(2.5 mol^(-2)))/((2 L)^(2))` `33.61 atm-5.61 atm=28.0` If the gas were considered is ideal gas, applying ideal gas equation, PV=nRT, we get `P=(nRT)/(V)=((2.5 mol)(0.0821" L atm mol"^(-1))(310 K))/(2 L)=31.8 atm` |
|
| 5. |
Calculate the pressure exerted by 0.350 moles of carbon dioxide in 0.360 L container at 100^@C. What pressure will be predicted by ideal gas equation? (a = 3.59 " atm " L^2 mol^(-2), b = 0.0427 L mol^(-1)) |
| Answer» SOLUTION :27.7 ATM , 29.8 atm | |
| 6. |
Calculate the precent dissociation ofH_(2) S (g) " if " 0*1 " mole of " H_(2) S " is kept in " 0*4 " litre vessel at 1000 K . For the reaction . " 2H_(2) S (g) hArr 2 H_(2) (g) + S_(2) (g), " the value of " k_(c) " is " 1*0 xx 10^(-6) |
|
Answer» Solution :`H_(2) S = (0*1 )/(0*4) "mol"L^(-1)= 0*25"mol" L^(-1)` Suppose degree of dissociation of `H_(2) S = alpha `. Then ` {: (,2 H_(2)S ,hArr,2H_(2),+,S_(2)),(" Intial conc.",0*25 M,,,,),("Conc.at eqm.",0*25 (1-alpha),,0*25 alpha = 0* 125 alpha ,,(0*25)/2alpha=0*125alpha):}` `K_(c) = ([H_(2)]^(2) [S_(2)])/([H_(2)S]^(2)):. 10^(-6) = ((0*25 alpha)^(2) (0*125 alpha))/[ 0*25 (1-alpha)]^(2)` Neglecting `alpha` in comparison to 1 , we get` 10^(-6) = ((0*25 alpha)^(2) ( 0*125 alpha))/(0*25 )^(2) ` ` 10^(-6)= 0*125 alpha^(3) or alpha^(3) = 8 xx 10^(-6)or alpha = 2 xx 10^(2) = 0*02 ` ` :. % " age dissociation "= 0*02 xx 100= 2 % ` |
|
| 7. |
Calculate the pressure exerted by 0.250 moles of carbon dioxide in 0.275 litres at 100^@C and compare this value with that expected for an ideal gas. (Given : a = 3.59 L^2 " atm " mol^(-2), b = 0.0427 L mol^(-1)) |
|
Answer» `(P+(an^2)/V_2)(V - nb) = nRT` Substituting the corresponding values, we have `[P+(3.59 xx (0.250)^2)/((0.275)^2)][0.275 - (0.250 xx 0.0427)]` `= 0.250 xx 0.0821 xx 373` which gives, P = 26 atm. Under similar conditions, for an IDEAL gas `P=(nRT)/V = (0.250 xx 0.0821 xx 375)/(0.275) =27.8` atm, Hence, the PRESSURE exerted by `CO_2` under given conditions is 26 atm. An ideal gas, under similar conditions will EXERT a pressure of 27.8 atm. |
|
| 8. |
Calculate the pOH of a solution obtained when 0.05 mol NH_(4)Cl is added and dissolved in 0.025 M ammonia solution. Kb for ammonia is 1.77xx10^(-5) |
|
Answer» SOLUTION :`pKb=log[KB]` `=log[1.77xx10^(-3)]` `=5-0.2480=4.752` `pOH=pK_(b)+log=(["salt"])/(["BASE"])=4.752+"log"([0.05])/([0.025])` `=4.752+log2=4.752+0.3010=5.053` |
|
| 9. |
Calculate the pOH of a solution at 25^@C that contains 1 xx 10^(-10) M of hydronium ions, i.e. H_3O^+. |
|
Answer» `4.000` At `25^@ C [H_3O^+][OH^-]=10^(-14)` `THEREFORE [OH^-]=10^(-14)/10^(-10)=10^(-4)` Now, `[OH^-]=10^(-p^"OH") =10^(-p^"OH")` `therefore` pOH=4 |
|
| 10. |
Calculate the pH which the following conversion will be at equilibrium in basic medium. I_(2)(s)hArrI^(-)(aq)+IO_(3)^(-)(aq) When the equilibrium concentrations at 300 K are [I^(-)]=0.1M and [IO_(3)^(-)]=0.1M Given that : DeltaG^(@)f (I^(-),aq)=-50 kJ mol ^(-1) DeltaG^(@)f (IO_(3),aq)=-123.kJ mol^(-1) DeltaG^(@)f (H_2O,l) =-233kJmol^(-1) DeltaG^(@)f (HO^(-),aq)=-150kJmol^(-1) R=(25)/(3)JK^(-1)mol^(-1) and ln 10 = 2.3 |
|
Answer» |
|
| 11. |
Calculate the pH value of (i) 10^(-2) "molar" HNO_(3) " solution" (ii)" 0.03 N HCl solution " (iii) 0.001 NH_(2)SO_(4) "solution ". |
|
Answer» Solution :(i) `HNO_(3)` completely ionizes as : `HNO_(3)+H_(2)O rarr H_(3)O^(+)+NO_(3)^(-)` `:. [H_(3)O^(+)]=10^(-2)M` (Given) `pH=-log[H_(3)O^(+)]=-log(10^(-2))=-(-2 log 10) = 2` (ii) HCl completely ionizes as : `HCl + H_(2)O rarr H_(3)O^(+) + Cl^(-)` `:. [H_(3)O^(+)]=[HCl]=0.03 N ` (Given) `=3xx10^(-2)N = 3xx10^(-2)M ` (HCl is monobasic, Eq. mass = Mol. mass) `:. pH = - log [H_(2)O^(+)]=- [log 3 xx10^(-2)]=-(log 3 + log 10^(-2))=- (0.4771-2) = 1.5229` (ii) `H_(2)SO_(4) ` completely ionizes as : `H_(2)SO_(4) + 2H_(2)O rarr 2H_(3)O^(+)+SO_(4)^(2-)` `{:(,[H_(3)O^(+)]=2xx[H_(2)SO_(4)],,[1 "MOLECULE of" H_(2)SO_(4) "gives"2H_(3)O^(+) "ions"]^(**),,),("But",H_(2)SO_(4)=0.001 N = 0.001 xx49 "g/litre",,("Eq. mass of " H_(2)SO_(4)=49),,),(,""=(0.001xx49)/(98) "moles/litre",,("Mol. mass of " H_(2)SO_(4)=98),,),(,""=0.0005 M ,,("DIRECTLY, Molarity"("Normality")/("Basicity")),,):}` `:. [H_(3)O^(+)]=2xx[H_(2)SO_(4)]=2xx0.0005M = 0.001 M = 10^(-3)M` `:. pH = - log [H_(3)O^(+)]=-log (10^(-3))=3` |
|
| 12. |
Calculate the pH value of a solution of 0.1 M NH_(3) (K_(b)=1.8xx10^(-5)) |
|
Answer» `:. [OH^(-)]=sqrt(K_(b)xx[NH_(3)]) `. Hence, `[H^(+)]=(K_(w))/([OH^(-)]), pH = - LOG [H^(+)]` Alternatively,`pOH = (1)/(2) [pK_(b)-log C]=(1)/(2) [-log (1.8xx10^(-5))-log (0.1)]=2.88` |
|
| 15. |
Calculate the pH value of (a) 0.0001 M NaOH (b) 0.01 M NaOH and (c) 0.04 M NaOH solutionat 25^(@)C. |
|
Answer» |
|
| 16. |
Calculate the pH value of a mixture containing 50 ml of 1 N HCl and 30 ml of 1 N NaOH solution, assuming both to be completely dissociated. |
|
Answer» |
|
| 17. |
Calculate the pH of the solution obtained by mixing 100 cm^(3) of solution with pH = 3with 400 cm^(3) of solution with pH = 4 . |
|
Answer» Solution :`100 cm^(3) ` of solution with PH = 3 CONTAINS `H^(+)=(10^(-3))/(1000)xx100=10^(-4)`mole `400cm^(3) ` of solution with pH = 4 contains `H^(+)=(10^(_4))/(1000)xx400=4XX10^(-5) ` mole TOTAL `H^(+)=10^(-4)+4xx10^(-5)=10^(-4)(1+0.4)=1.4xx10^(4)`. Total volume = `500 cm^(3)` `:. [H^(+)]=(1.4xx10^(-4))/(500) xx1000M = 2.8xx10^(-4)M` `pH = - log (2.8xx10^(-4))=4-0.4472 ~~ 3.55` |
|
| 18. |
Calculate the pH of the solution in which 0.2 M NH_4Cl and 0.1 M NH_3 are present . The pK_b of ammonia solution is pK_b= 4.75 |
|
Answer» Solution :Calculation of `K_b` : `pK_b=-log (K_b)`=(4.75) `therefore log K_b=-4.75` `therefore K_b=10^(-4.75)=1.7783xx10^(-5)` `approx 1.78xx10^(-5)` Calculation of `[OH^-]` : `{:("Strong salt :", NH_4Cl to , NH_(4(AQ))^(+)+, CL^(-)),(,0.2,0.2M,0.2M):}` `{:("Weak base: ",NH_(3(aq))+ , H_2O_((l)) hArr , NH_(4(aq))^(+)+, OH_((aq))^(-)),("Equilibrium of " NH_3 " Initial (M):", 0.1 , -, +0,+0),("Change reaction :", -X M, ,+xM,+xM),("CONCENTRATION at equilibrium:",(0.1-x)M,,(0.2+x)M,xM) :}` `K_b=([NH_4^+][OH^-])/([NH_3])` `1.78xx10^(-5)=((0.2+x)(x))/((0.1-x))` x is neglected in additional substraction . Value of `K_b` is much less , the value of x is very less , Therefore x is NEGLIGIBLE in comparison to 0.2 and 0.1 . `therefore (0.2+x) approx 0.2` and `(0.1 -x) approx 0.1` `therefore 1.78xx10^(-5) =(0.2x)/0.1` `therefore x=[OH^-]=(1.78xx10^(-5)xx0.1)/0.2` `= 0.88xx10^(-5)` M Calculationof `[H^+]` : `[H^+] =K_w/([OH^-]) = (1xx10^(-14))/(0.88xx10^(-5))=1.136xx10^(-9)` Calculation of pH : pH=-log `[H^+] = log 1.136xx10^(-9) = 8.9445 approx` 8.95 |
|
| 19. |
Calculate the pH of the resultant mixtures : 10 mL 0.1 M H_2SO_4 + 10 mL 0.1 M KOH |
|
Answer» SOLUTION :Millimol=mL x M In Initial MOL of `H_2SO_4` = 10 x 0.1 = 1.0 mol `H_2SO_4` In initial m mole of KOH = 10 x 0.1 = 1.0 m mol KOH Total volume of mix solution20 mL = 0.02 L Millimol of remaining `H_2SO_4` after neutralization `{:(H_2SO_4 + , 2KOH to , K_2SO_4 + , 2H_2O),("1 mol","2 mol" , "1 mol", "2 mol"):}` `therefore` (m mol of remaining `H_2SO_4` ) = (m mol `H_2SO_4` in initial ) -(`H_2SO_4` consumed in neutralization ) =(1 m mol- `1/2` m mol ) `H_2SO_4` =0.5 m mol `H_2SO_4` remaining =0.5/1000 mol `H_2SO_4` remaining Molarity of remaining , `H_2SO_4 =("mol"/"Total volume (L)")` `=(0.5xx1000)/(1000xx20)` `=0.5/20`=0.025 M `H_2SO_4` Remaining `[H^+]` in mixture = `2[H_2SO_4]` =2(0.025) = 0.05 PH = log `[H^+]` = log (0.05) =-(-1.3010)=1.3010 |
|
| 20. |
Calculate the pH of the resultant mixtures : 10 mL 0.2 M Ca(OH)_2+ 25 mL 0.1 M HCl |
|
Answer» Solution :MOLARITY =Mol/Litre, So, Mol=Molarity X Litre 10 mL 0.2 M `Ca(OH)_2=10/1000` L x 0.2 M `=2/1000 "mol" Ca(OH)_2` =2 millimole `Ca(OH)_2` 25 mL 0.1 M HCl = `25/1000` L x 0.1 M `=2.5/1000`mol HCl =2.5 millimole HCl REMAINING substrate of neutralization: `{:(Ca(OH)_2+, 2HCl to , CaCl_2+,2H_2O),("1 mol","2 mol","1 mol","2 mol"):}` Here, 2 mol HCl react with 1 mol `Ca(OH)_2` `therefore 2.5/1000` mol HCl `to 1.25/1000` mol `Ca(OH)_2` react. `therefore` The remaining mol of `Ca(OH)_2`after neutralisation. `(("Initial"),(2.0/1000))-(("After REACTION"),(1.25/1000))` `=0.75/1000` mol `Ca(OH)_2` is remainly Molarity of remaining `Ca(OH)_2=("mol"/"TOTAL volume (L)")` `=(0.75/1000)1000/35`=0.02143 M `[OH^-]=2[Ca(OH)_2]=2xx0.02143` =0.04286 M =`4.286xx10^(-2)` M pOH=-log `[OH^-]=-log (4.286xx10^(-2))` =-(-1.3678)=1.3679 pH=(14.0-pOH)=(14.0-1.3679) =12.632 =12.63 |
|
| 21. |
Calculate the pH of the resultant mixtures : 10 mL 0.01 M H_2SO_4 + 10 mL 0.01 M Ca(OH)_2 |
|
Answer» SOLUTION :10 mL 0.01 M `H_2SO_4 = (10xx0.01)` =0.1 millimol `H_2SO_4` 10 mL 0.01 M `Ca(OH)_2=(10xx0.01)` = 0.1 millimol `Ca(OH)_2` Total volume 20 ml = `20/1000` L = 0.02 L NeutralisationReaction : `{:(Ca(OH)_2 +,H_2SO_4 to , CaSO_4 + , 2H_2O),("1 MOL", "1 mol" , " 1 mol" , "2 mol") :}` 0.1 m mol + 0.1 m mol `to` 0.1 m mol Thus, `H_2SO_4` and `Ca(OH)_2` COMPLETE neutralise mixture remain neutral. `therefore` pH=7.0 |
|
| 22. |
Calculate the pH of the following solutions : (a) 2 g of TlOH dissolved in water to give 2 litre of the solution (b) 0.3 g of Ca(OH)_(2) dissolved in water to give 500 mL of the solution (c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution (d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of the solution. |
|
Answer» Solution :(a) Molar conc. Of TLOH `= (2G)/((204+16+1)g "MOL"^(-1))xx(1)/(2L)=4.52xx10^(-3)M` `:. [OH^(-)]=[TlOH]=4.52xx10^(-3)M` `[H^(+)]=10^(-14)//(4.52xx10^(-3))=2.21xx10^(-12)M` `pH = - log (2.21xx10^(-12))=12-(0.3424)=11.66` (b) Molar conc. of `CA(OH)_(2) = (0.3 g)/((40+34) g "mol"^(-1))xx(1)/(0.5L)=8.11xx10^(-3)M` `Ca(OH)_(2) rarr Ca^(2+)+ 2 OH^(-)` `:. [OH^(-)]=2[Ca(OH)_(2)]=2xx(8.11xx10^(-3))M = 16.22xx10^(-3)M` `pOH = - log (16.22xx10^(-3))=3-1.2101=1.79` `pH = 14-1.79=12.21` (C) Molar conc. of NaOH `=(0.3 g)/(40 g "mol"^(-1))xx(0.2L)=3.75xx10^(-2)M` `[OH^(-)]=3.75xx10^(-2)M` `pOH = - log (3.75xx10^(-2))=2-0.0574 = 1.43 :. pH = 14 - 1.43 = 12.57` (d) `M_(1)V_(1) = M_(2)V_(2) :. 13.6 M xx 1 mL = M_(2) xx 1000mL :. M_(2) = 1.36xx10^(-2)M` `[H^(+)]=[HCl]=1.36xx10^(-2)M, pH = - log (1.36xx10^(-2))=2-0.1335 ~= 1.87` |
|
| 23. |
Calculate the pH of the following solutions : (a) 2 g of TIOH dissolved in water to give 2 litre of solution. (b) 0.3 g of Ca(OH)_2 dissolved in water to give 500 ml, of solution. (c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. (d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution. |
|
Answer» SOLUTION :(a) pH of 2 L solution of 2g TlOH : Molecular mass of TlOH =204+16+1=221 g mol Molarity of TlOH =`"Weight"/"Molecular mass x Litre"` `=2/(222xx2)=4.525xx10^(-3)` M TlOH is strong base, complete ionization occurs so in solution, `[OH^-]=[TlOH] = 4.525xx10^(-3)` M pOH=-log `(4.525xx10^(-3))` `=-[log 4.525+log 10^(-3)]` =-(0.6556-3.0)=2.3444 pH = 14.0 - pOH = 14.0 - 2.3444 = 11.6556 `approx` 11.66 (b) pH of 500 ML solution of 0.3 g `Ca(OH)_2` : Molecular mass of `Ca(OH)_2`=40+(16+1)2 = 74 g `mol^(-1)` `M="Weight in g"/"Molecular mass x VOLUME"` `=((0.3g))/((74 "g mol"^(-1))(0.5L))=8.108xx10^(-3)` M `Ca(OH)_2` is strong , 100% DISSOCIATION of base, `Ca(OH)_2 to Ca_((aq))^(2+) + 2OH_((aq))^(-)` `therefore` In solution `[OH^-]=2xx[Ca(OH)_2]` `=(2xx8.108xx10^(-3))M` `=16.216xx10^(-3)` M pOH=log `[OH^-]`=-log `(16.216xx10^(-3))` =-(1.2099-3.0)=-(-1.7901) =+1.7901 pH = 14.0 -pOH =14.0-1.7901=12.2099 `approx`12.21 (c) pH of 200 mL solution of 0.3 g NaOH : Molarity = `"Weight x 1000"/"Molecular mass x Volume mL"` `=(0.3xx1000)/(40xx200)=(0.3 g xx 1000)/((40 g "mol"^(-1))xx(200 mL))` `=3.75xx10^(-2)` M Strong base NaOH complete ionization in solution, `[OH^-]=[NaOH ]=3.75xx10^(-2)` M `pOH=-log [OH^-]=-log (3.75xx10^(-2))` =-(0.5740-2.0)=-(-1.426)=1.426 pH = 14.0-pOH =14.0 - 1.426 = 12.574 `approx` 12.57 (d) pH of 1 L solution MADE in 1 mL of 13.6 M HCl : `{:("Initial stage","After dilution"),(M_1V_1=,M_2V_2):}` `therefore 13.6xx1 mL = M_2xx1000 mL` `therefore M_2=1.36xx10^(-2) M HCl =[H^+]` `pH = -log [H^+]- - log (1.36xx10^(-2))` =2-0.1335= 1.8665 `approx` 1.87 |
|
| 24. |
Calculate the pH of the following resultant mixtures : (a) 10 mL of 0.2 M Ca(OH)_(2) + 25 mL of 0.1 M HCl (b) 10 mL of 0.01 M H_(2)SO_(4) + 10 mL of 0.01 M Ca(OH)_(2) (c) 10 mL of 0.1 M H_(2)SO_(4) + 10mL of 0.1 M KOH. |
|
Answer» SOLUTION :(a) mL of 0.2 M `Ca(OH)_(2)=10xx0.2` MILLIMOLES = 2 millimoles of `Ca(OH)_(2)` 25mL of 0.1 M HCL `= 25xx0.1` millimoles = 2.5 millimoles ofHCl `Ca(OH)_(2)+2HCl rarr CaCl_(2)+2 H_(2)O` 1 millimole of `Ca(OH)_(2) + 2 HCl rarr CaCl_(2) + 2H_(2)O` 1 millimole of `Ca(OH)_(2)` reacts with 2 millimoles of HCl `:.` 2.5 millimoles of HCl will react with 1.25 millimoles of `Ca(OH)_(2)` `:. Ca(OH)_(2) ` left `= 2-1.25 = 0.75` millimoles(HCl is the limiting reactant) Total volume of the solution = 10 + 25 mL = 35 mL `:.` Molarity of `Ca(OH)_(2)` in the mixture solution `=(0.75)/(35) M = 0.0214M` `:. [OH^(-)]=2xx0.0214 M = 0.0428 M = 4.28 xx 10^(-2)` `pOH = - log (4.28xx10^(-2))=2-0.6314=1.39=686~=1.37` `:. pH = 14 - 1.37 = 12.63` (b) 10 mL of 0.01 M `H_(2)SO_(4) = 10xx0.01` millimole = 0.1 millimole 10 mL of 0.01 M `Ca(OH)_(2) = 10 xx 0.01` millimole = 0.1 millimole `Ca(OH)_(2) + H_(2)SO_(4) rarr CaSO_(4) + 2H_(2)O` 1 mole of `Ca(OH)_(2) ` reacts with 1 mole of `H_(2)SO_(4)` ltbvrgt `:. ` 0.1 millimole of `Ca(OH)_(2)` will react completely with 0.1 millimole of `H_(2)SO_(4)`. Hence, solution will be neutral with pH = 7.0 (c) 10 mL of 0.1 M `H_(2)SO_(4) = 1 ` millimole 10 mL of 0.1 M KOH = 1 millimole `2KOH + H_(2)SO_(4) rarr K_(2)SO_(4) + 2 H_(2)O` 1 millimole of KOH will react with 0.5 millimole of `H_(2)SO_(4)` `:. H_(2)SO_(4) ` left = 1 - 0.5 = 0.5 millimole Volume of reaction mixture = 10 + 10 = 20 mL `:. ` Molarity of `H_(2)SO_(4)` in the mixture solution `=(0.5)/(20) = 2.5xx106(-12) M` `[H^(+)]=2xx(2.5XX10^(-2))=5xx10^(-2)` `pH = - log (5xx10^(-2))=2-0.699=1.3` |
|
| 25. |
Calculate the pH of the following mixtires of strong acids, strong bases, and combination of both: a. 500mL of 0.1M HCI + 200 mL of 0.1 M H_(2)SO_(4) + 300 mL of 0.2 M HNO_(3) b. 100mL of 0.1 M HCI +100 mL of 0.2 M H_(2)SO_(4) + 100 mL of 0.1 M HNO_(3) and 700 mL of H_(2)O c. 500mL of 0.1 M NaOH +100 mL of 0.1 M Ca(OH)_(2) +400 mL of 0.2 M KOH d. 100mL of 0.1 M Na OH +200mL of 0.1 N Ca (OH)_(2) +200mL of 0.1 M KOH and 500mL of H_(2)O e. 100mL of 0.1M HCI +300mL of 0.1 M H_(2)SO_(4) +100 mL of 0.3 M Ba (OH)_(2) and volume was made to 1L by adding water f 500 mL of 0.1 M HCI +100 mL of 0.1 N H_(2)SO_(4)+ 400 mL of 0.1 M Ca (OH)_(2) g 8g of NaOH +680mL of M HCI +10 mL of H_(2)SO_(4), (specific gravity 1.2, 49% H_(2)SO_(4) bu mass). The total volume of the solution was made to 1L with water. h. 37.0 g of Ca(OH)_(2) +360mL of 1M HCI +10 mL of H_(2)SO_(4) (density = 1.4, 49% H_(2)SO_(4) by mass). The total volume of the solution was made to 1L with water. |
|
Answer» Solution :a. All are strong acid and completely ionised. Total volume `= 500 +200 +300 = 100 mL` `N_(1)V_(1) +N_(2)V_(2) +N_(3)V_(3) +N_(4)V_(4)` `N_(4) = (N_(1)V_(1) +N_(2)V_(2) +N_(3)V_(3))/((V_(1)+V_(2)+V_(3))` `(V_(4) = V_(1) +V_(2) +V_(3) = 1000mL)` `0.1 xx 1 xx 500 xx 0.1(n` factor) `xx 200` `N_(A) = (+0.2 xx 1xxx 300)/(1000)` `= (50 +40 +60)/(1000) = (150)/(1000) = 15 xx 10^(-2)N` `PH =- log [H_(3)O^(o+)]` `=- log (15 xx 10^(-2))` `=- log (5 xx3) - log (10^(-2))` `=- 0.7 - 0.48 +2 = 0.82` b. Total volume `= 100 +100 +100 +700 mL H_(2)O` `= 1000mL` `N_(A)=(100xx0.1xx1+100xx0.2xx2(n"factor")+100xx0.1xx1)/(1000)` `= (10+40 +10)/(1000) = (60)/(1000) = 6 xx 10^(-2)N` `:. pH =- log (6 xx 10^(-2)) =- log (3xx2) - log 10^(-2)` `=- log 3 - log 2 +2` `= - 0.48 - 0.3 +2 = 1.22` c. All are strong bases and completely ionised. Thus first claculate `[overset(Θ)OH]` and `pOH` than calculate `pH` accordingly `N_(4)=(500 xx 0.1 xx 1 +100 xx 0.1 xx 2(n "factor") +400 xx 0.2 xx 1)/(500+100+400)` `= (50+20+80)/(1000) = (150)/(1000) = 15 xx 10^(-2) N` `:. pOH =- log [overset(Θ)OH] =- log (15 xx 10^(-2))` `=- log (5 xx 3) - log 10^(-2)` ` =- log 5 - log 3+2` `=- 0.7 - 0.48 +2` `= 0.82` `pH = 14 - 0.82 = 13.18` d. Note: Solution of `CA(OH)_(2)` is given in normally so NEED not to be multiplied by 'n factor' `N_(4) = (100 xx 0.1 xx 1 +200 xx 0.1 xx1 +200 xx 0.1 xx)/(100+200+200 +500 "mL of" H_(2)O)` `= (10 +20 +20)/(1000) = (50)/(1000) = 5 xx 10^(-2)N` `:. pOH =- log [overset(Θ)OH] =- log (5 xx 10^(-2))` `=- log 5 - log 10^(-2)` `=- 0.7 +2 = 1.3` `:. pH = 14 - 1.3 = 12.7` e. Here, mixture of strong acid and strong bases is given. So first calculate th `mEq` of acids and bases. i. IF the `mEq` of strong acid is in excess, then calculate the `pH` from the concentration of `[H^(o+)]` left. ii. If the `mEq` of strong bases is in excess, then calculate the `pOH` from the concentration of `[overset(Θ)OH]` left and then calculate `pH` accrodingly. Total `mEq` of strong acids `= 100 xx 0.1 xx 1 +300 xx 0.1 xx 1("n factor")` `= 10 +60 = 70 mEq` Total `mEq` strong base `= 100 xx 0.3 xx 2("n factor")` `= 60 mEq` `mEq` of strong acid left `= 70 - 60 = 10 mEq` Total volume of solution `= 1L = 1000 mL` `:. [H^(o+)] = (10)/(1000) =10^(-2)N` `pH =- log (10^(-2)) = 2` f. Proceed as in part (e). (Here `H_(2)So_(4)` is given in normality) Total `mEq` of strong acid `= 500 xx 0.1 xx 1 +100 xx 0.1 xx1` `= 50 +10 =60 mEq` Total `mEq` of storng base `= 400 xx 0.1 xx 2("n factor")` `= 8- mEq` `mEq` of strong base left =` 80 - 60 = 20 mEq` Total volume of solution `= 500 +100 +400` `1000 mL` `:.[overset(Θ)OH] = (20)/(100) = 2 xx 10^(-2)N` `:. pOh = - log (2xx 10^(-2))` `=- log2 - log 10^(-2)` `=- 0.3 +2 = 1.7` Thus, `pH = 14- 1.7 = 12.3`. g. Moles of `NaOH = (8)/(40) (Mw of NaOH = 40)` `0.2 mol = 200 m mol` `N_(H_(2)SO_(4)=(%"by weight" xx 10 xxd)/(Ew_(2))` `= (49 xx 10 xx 1.2)/(49) = 12N` `mEq` of `H_(2)SO_(4) = 12N xx 10mL = 120 mEq` `mEq` of `HCI = 680 xx 1 xx 1 = 680 mEq` Total `mEq` of acid `= 120 +680 = 800 mEq` `mEq` of acid left `= 800 - 200 = 600 mEq` Total volume `= 1000 mL` `:. [H^(o+)] = (600)/(1000) = 6 xx 10^(-1)N` `pH =- log (6 xx 10^(-1)) =- log (3xx2) - log 10^(-1)` `=- log 3 - log2 +1` `=- 0.48 - 0.3 +1 = 0.22` h. Mw of `Ca(OH)_(2) = 74` Ew of `Ca (OH)_(2) = (74)/(2) = 37 g Eq^(-1)` eq of `ca (OH)_(2) = (37)/(37) = 1 Eq = 1000 mEq` `N_(H_(2)SO_(4)) = (% "by weight" xx 10 xx d)/(Ew_(2)) = (49 xx 10 xx 1.4)/(49)` `= 14N` `mEq` of `H_(2)SO_(4) = 14N xx 10mL = 140meq` `mEq` of `HCI = 360 mL xx 1N = 360 mEq` Total `mEq` of acid`= (140 +360) = 500 mEq` `mEq` of base left `= 100 - 500 = 500 mEq` Total volume `= 1000 mL` `:. [overset(Θ)OH] (500)/(1000) = 0.5N = 5 xx 10^(-1)N` `pOH =- log [overset(Θ)OH]` `=- log (5 xx 10^(-1))` `=- log 5 - log 10^(-1) =- 0.7 +1 = 0.3` Thus, `pH = 14 - 0.3 = 13.7` |
|
| 26. |
Calculate the pH of solution prepared by mixing 40 ml of 0.1 Macetic acid with 40 ml of 0.1 M sodiumacetate, Given K_a = 10 ^(-5) |
|
Answer» ` PH =pKa + LOG ""(S)/(A)= pKa = 5` |
|
| 27. |
Calculate the p of solution obtained by mixign 10 ml fo 0.1 M HCl and 40 ml off 0.2 M H_(2)SO_(4). |
|
Answer» ` PH =PKA + log ""(S)/(A)= pKa = 5` |
|
| 28. |
Calculate the pH of solution made by mixing equal volume of : a. Two solutions having pH = 1.5 and 2.5. b. Three solutions having pH = 15,2.5, and 3.5. c. Two solutions having pH = 8 and 9. d. Three solutions having pH 8,9, and 10. e. Two solutions having pH =2and 4. f.Three solutions having pH = 2,4, and 6. |
|
Answer» Solution :a. The two `pH` values differe by one unit. Approximate final `pH = ((1.5 +2.5))/(2) = 2` Actural final `pH = 2.0 - 0.24 = 1.76` Checking final pH : `pH = 1.5, log [H^(o+)] =- 1.5 =- 1- 0.5 +1 -1 = bar(2).5` `[H^(o+)] = "Antilog" (bar(2).5) ~~3xx 10^(-2)N` `pH = 2.5, log [H^(o)]=- 2.5 =p- 2- 0.5 +1 -1 = bar(3).5` `[H^(o+)] = "Antilog" (bar(3).5) ~~ 3xx 10^(-3)N` Final `[H^(o+)] = ((3xx10^(-2)+3xx10^(-3)))/(2)` `= (3xx10^(-2))/(2) [1+0.1]` `=(1.1)/(2) xx3xx 10^(-2)` `0.55 xx 3xx 10^(-2) = 55 xx 3xx10^(-4)` `pH =- log 3 - log 10^(-4)` `=- 1.74 - 0.5 +4` (taking log `3~~0.48 ~~0.5`) `=-2.24 +4 = 1.76` b. The three `pH` values differe by one unit Approximate final `= ((1.5+2.5+3.5))/(3) = (7.5)/(3) = 2.5` Actual final `pH = 2.5 - 0.56 = 1.94` Checking final pH : Proceed as above in pair (a) `pH = 1.5, :. [H^(o+)] ~~3xx10^(-2)N` `pH =2.5, :. [H^(o+)] ~~3 xx 10^(-3)N` `pH = 3.5 :. [H^(o+)]~~3 xx 10^(-4)N` Final `[H^(o+)] = (3xx10^(-2)+10^(-3)+10^(-4))/(3)` `=3xx10^(-2) (1+0.1+0.01)` (neglect 0.01) `= (1.1 xx 3)/(3) xx 10^(-2)` `=0.366 xx3x10^(-2) = 366 xx 3xx 10^(-5)` `pH =- log (366) - log3 - log 10^(-5)` `=- 2.5635 - 0.5 +5` (taking log `3 = 0.48 ~~0.5`) `=-3.0635 +5 = 1.9365 ~~1.94` `pH = 1.94` c. Although the `pH` values are in basic range, so `pH` of mixture can be calculated as when `pH` values are in acidic range. Approximate final `pH = ((8+9))/(2) = 8.5` Actual final `pH = 8.5 - 0.24 = 8.26` Checking final pH : `pH = 8:. [H^(o+)] = 10^(-8)N` `pH = 9:. [H^(o+)] = 10^(-9)N` Final `[H^(o+)] = ((10^(-8)+10^(-9))/(2)` `=(10^(-8)(1+0.1))/(2)` `= (1.1)/(2) xx 10^(-8)N` `=0.55 xx 10^(-8)= 55 xx 10^(-10)N` `pH =- log (55) - log10^(-10) =- 1.74 +100 = 8.26`. d. Procedd as in part (c) Approximate final `pH = ((8+9+10))/(3) = 9` Actual final `pH = 9.0 - 0.56 = 8.44` Checking final pH : Final `[H^(o+)] = (10^(-8)+10^(-9)+10^(-10))/(3)` `= (10^(-8)(1+0.1+0.01))/(3)` (neglect 0.01) `= (1.1)/(3) xx 10^(-8) = 0.366 xx 10^(-8) = 366 xx 10^(-11)` `pH =- log (366) - log 10^(-11)` `=- 2.5635 +11 = 8.365 ~~ 8.44` e. The two `pH` values do not differ by one unit, so calulate the `pH` directly `[H^(o+)]_("TOTAL") = ((10^(-2)+10^(-4))/(2)) = (10^(-2))/(2) (1+10^(-2))` (neglect `10^(-2)`) `~~(10^(-2))/(2) ~~0.5 xx 10^(-2)` `~~5xx10^(-3)M` `pH ~~ - log (5x10^(-3)) ~~- 0.7 +3 ~~2.7` f. Here also, the three `pH` VALUE do not differ by one unit, so calculate the `pH` directly. `[H^(o+)]_("total") = ((10^(-2)+10^(-4)+10^(-6))/(3)) = (10^(-2))/(3) (1+10^(-2)+10^(-4))` (neglect `10^(-2)` and `10^(-4)` ) `~~ (10^(-2))/(3)~~ 0.3 xx 10^(-2)` `~~ 3 xx 10^(-3)M` `pH ~~ - log (3xx10^(-3)) ~~ 0.48 +3 ~~ 2.52` |
|
| 29. |
Calculate the pH of N/1000 sodium hydroxide solution assuming complete ionization (K_(w)=1.0xx10^(-14)) |
|
Answer» Solution :NaOH completely IONIZES as : `NaOH rarr Na^(+) + OH^(-)` `:. [OH^(-)]=[NaOH]=(N)/(1000)=10^(-3)N=10^(-3)M ` (`:'` Eq. mass = Mol. Mass in case of NaOH) Now, as `[H_(3)O^(+)][OH^(-)]=10^(-14) :. [H_(3)O^(+)]=(10^(-14))/([OH^(-)])=(10^(-14))/(10^(-3))=10^(-11)` `:. PH = - log [H_(3)O^(+)]= - log 10^(-11) =11` |
|
| 30. |
Calculate the pH of one molar ammonium formate solution. Given that pK_a for HCOOH = 3.8,PK_b for NH_3 = 4.8. |
|
Answer» Solution :HYDROLYSIS of AMMONIUM acetate is given by the equation as `HCOONH_4 +H_2O hArr HCOOH +NH_4 OH` PH of the solution of a salt of weak ACID and weak base `=1/2[pK_w+PK_a-PK_b ]` pH of 1M `HCOONH_4` solution`= 1/2 (14+3.8-4.8) = 6.5`. |
|
| 31. |
Calculate the pH of (i) 1 M H_(2)SO_(4) " " (ii) 2 M H_(2)SO_(4) " " (iii) 10^(-2)MH_(2)SO_(4) solutions. Given that the second ionization constant (K_(a_(2))) " of"H_(2)SO_(4) "if" 10^(-2). |
|
Answer» Solution :`H_(2)SO_(4)` ionizes in two steps as `H_(2)SO_(4) rarr H^(+)+HSO_(4)^(-) ` (almost completely ionized ) `HSO_(4)^(-) hArr H^(+) + SO_(4)^(2-)` (partially ionized) (i) In 1M `H_(2)SO_(4), H^(+)` ions are obtained MAINLY from 1st ionization = 1 M and `H^(+)` ions obtained from 2nd STEP are negligible as compared to 1 M. Hence `[H^(+)] = 1 M , pH = - log 1 = 0 `. (ii) In 2 M `H_(2)SO_(4)` , again `H^(+)` ions are obtained mainly from 1st ionization as explained above. Hence, `[H^(+)]=2 M, pH = - log 2 = - 0.303`. Here note that there is nothing wrong with the negative value of pH but this offers no advantage over the actual value of `[H^(+)]`. (iii) In `10^(-2) M H_(2)SO_(4),H^(+)` ions obtainedfrom2nd ionization cannot be neglected in comparison to `10^(-2) M H^(+)` ions obtained from 1 st ionization. Hence, we have `{:(,H_(2)SO_(4),rarr,H^(+),+,HSO_(4)^(-),),("INITIAL",10^(-2)M,,,,,),("After1 st ionization",~=0,,10^(-2)M,,10^(-2)M,),("Initial",HSO_(4)^(-),hArr,H^(+),+,SO_(4)^(-),),(,10^(-2)M,,,,,),("After ionization",10^(-2)-x,,10^(-2)+x,,x,):} (H^(+)=10^(-2) "from 1 st ionization + x from 2nd ionization")` `:. K_(a_(2))=([H^(+)][SO_(4)^(2-)])/([HSO_(4)^(-)])=((10^(-2)+x)(x))/(10^(-2)-x)=10^(-2)` (Given) `:. 10^(-2)x + x^(2)=10^(-4)-10^(-2)x or x^(2) + 2 XX 10^(-2)x-10^(-4) = 0 ` `:. x=(-b pmsqrt(b^(2)-4AC))/(2a) = (-2xx10^(-2)pmsqrt(4xx10^(-4)+4xx10^(-4)))/(2) = (-2xx10^(-2)+2.828xx10^(-2))/(2)` `=(0.828xx10^(-2))/(2)` (ignoring the -ve value ) `=0.414xx10^(-2) M` `:. [H^(+)]=10^(-2)+x=10^(-2)+0.414xx10^(-2) = 1.414xx10^(-2)` `:. pH = - log [H^(+)]=-log(1.414xx10^(-2))=1.85` |
|
| 32. |
Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation (pK_(a) of formic acid = 3.8 and pK_(b) of ammonia = 4.8 ). |
|
Answer» |
|
| 33. |
Calculate the pH of a solution whose Hydroge ion conc.is 1.5xx10^(3)M. |
|
Answer» SOLUTION :`pH=-LOG[H^(+)]` `=-log(1.5xx10^(-3))` `=3-log1.5=3-0.1761` `=2.8239` |
|
| 34. |
Calculate the pH of a solution which is 1xx10^(-3) M with respect to sulphuric acid. |
|
Answer» |
|
| 35. |
Calculate the pH of a solution that contains 1.00 M HF (K_(a)=7.2xx10^(-4)) and 5.00M HClO (K_(a)=3.5xx10^(-8)). |
|
Answer» Solution :As `K_(a)` for `HClO lt ltK_(a)` for HF, `[H_(3)O^(+)]` obtained from HClO can be neglected in comparison to `[H_(3)O^(+)]` obtained from HF. HENCE `[H_(3)O^(+)]=SQRT(K_(a) (HF)xxC(HF))=sqrt((7.2xx10^(-4))(1.0))=2.68xx10^(-2)M` `pH = - log (2.68 xx 10^(-2))=2-0.4265 = 1.57` (approx) |
|
| 36. |
Calculate the pH of a solution obtained by mixing equal volumesof thesolutions withpH = 3 and pH = 5. |
|
Answer» Solution :pH = 3 MEANS `[H^(+)]=10^(-3)M and pH = 5` means `[H^(+)]=10^(-5)`M. On mixing equal volumes, inthe final solution, `[H^(+)]=((10^(-3)+10^(-5))/(2)) = (10^(-3)(1+10^(-2)))/(2) = (1.01xx10^(-3))/(2) = 0.505xx10^(-3) M = 5.05 xx 10^(-4) M ` `:. pH = - log (5.05xx10^(-4))=4-0.7033~~3.3` |
|
| 37. |
Calculate the pH of a solution obtained by mixingequal volumes of N/10 NaOH and N/20 HCl. |
|
Answer» 1 L of N/10 NaOH containsNaOH `=(1)/(10)" g eq." = 0.1 g eq`. 1 L of N/20 HCl contains HCl `=(1)/(20) g eq. = 0.05 g eq`. 0.05 g eq. of HCl will neutralise 0.05 g eq. of NaOH. `:. ` NaOH left unneutralised = 0.05 g eq. Volume of solution = 2 L `:. `Concentration of NaOH in the FINAL solution `= (0.05)/(2)" g eq" L^(-1) = 0.025 N` `[NaOH]=[OH^(-)] = 0.025 M = 2.5 xx 10^(-2) :. [H_(3)O^(+)]=(10^(-14))/(2.5xx10^(-2))=4xx10^(-13) M` |
|
| 38. |
Calculate the pH of a solution obtained by mixing 50 ml of 0.2 M HCl with 49.9 ml of 0.2 M NaOH solution. |
|
Answer» Hence, `pH = - log (2xx10^(-4))=4-0.301=3.699` |
|
| 39. |
Calculate the pH of a solution obtained by mixing 5 mL of 0.1 M NH_(4) OH with 250 mL of 0.1 M NH_(4) Cl solution . K_(b) for NH_(4)OH=1.8xx10^(-5). |
|
Answer» Solution :5 mL of 0.1 M `NH_(4)OH = 5 xx 0.1` MILLIMOLE = 0.5 millimole 250 mL of 0.1 M `NH_(4)Cl= 250 xx 0.1` millimole = 25 MILLIMOLES Total VOLUME of solution after mixing = 255 mL `:. `[Salt]=`[NH_(4)Cl] = (25)/(255) M` [Base]`=[NH_(4)OH]=(0.5)/(255) M` `POH = pK_(B) + log. (["Salt"])/(["Base"])=-log. (1.8xx10^(-5))+log.(25//255)/(0.5//255) = (5-0.2553) + 1.6990= 6.4437` `pH = 14-6.4437 = 7.5563` |
|
| 40. |
Calculate the pH of a solution obtained by diluting 25 ml of N/100 HCl to 500 ml. |
|
Answer» `[H^(+)]=[HCL] = 5xx10^(-4)M` `pH = - LOG (5xx10^(-4))=4-log 5 = 4 - 0.6990 = 3.301`. |
|
| 41. |
Calculate the pH of a solution made by mixing 0.1 M NH_(3) and 0.1M (NH_(4))_(2)SO_(4). (pK_(b) of NH_(3) = 4.76) |
|
Answer» Solution :It is a BASIC BUFFER. `["Base"] = [NH_(3)] = 0.1m` `["Salt"] = [(NH_(4))_(2)SO_(4)] = 0.1 xx2 = 0.2M` `[(NH_(4))_(2) SO_(4) rarr 2overset(o+)NH_(4) +SO_(4)^(2-)]` `pOH = pK_(B) + "log"(["Salt"])/(["Base"])` `4.76 +log (0.2M)/(0.1M)` `= 4.76 +LOG2` `= 4.76 +0.3 = 5.03` `:. pH = 14 - 5.03 = 8.97` |
|
| 42. |
Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH =4 respectively. |
|
Answer» Solution :pH of solution A = 6. Hence, `[H^+] = 10^(-6) "mol L"^(-1)` pH of solution B = 4. Hence, `[H^+]10^(-4) "mol L"^(-1)` On mixing 1 L of each solution, MOLAR concentration of total `H^+` is halved. Total, `[H^+]= (10^(-6) + 10^(-4))/2 "mol L"^(-1)` `[H^+]=(0.01+10^(-4))/2 5.05 XX 10^(-5) "mol L"^(-1)` `[H^+] = 5.0xx10^(-5) "mol L"^(-1)` `pH=-log [H^+] rArr pH-log (5.0xx10^(-5))` `pH=-[log 5+ (-5 log 10 )] rArr pH =- log 5+5` `pH=5-log 5 =5-0.6990 rArr pH` 4.3010 = 4.3 Thus, the pH of RESULTING solution is 4.3 |
|
| 43. |
Calculate the pH of a solution formed on mixing 0.2 M NH_(4)Cl and 0.1 M NH_(3). The pK_(b) of ammonia solution is 4.75. |
|
Answer» Solution :`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-), pK_(b)=4.75` On mixing `NH_(3) ` with `NH_(4)Cl` solution, we have `{:(,NH_(3),+,H_(2)O,hArr,NH_(4)^(+),+,OH^(-)),("Initial conc.",0.1M,,,,0.2M,,0),("At eqm.",0.1 -x,,,,0.2+x,,x):}` `K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` As`pK_(b) = 4.75 , i.e., -log K_(b) = 4.75 or log K_(b) = -4.75 = bar(5) .25 :. K_(b) = 1.77xx10^(-5)` `:. 1.77xx10^(-5) = ((0.2+x)(x))/(0.1-x)~=((0.2)x)/(0.1)` (neglecting x in comparison to 0.1 and 0.2) or,`x=010^(-5),i..e., [OH^(-)]=0.885xx10^(-5)=8.85xx10^(-6)M` or `[H^(+)]=(10^(-14))/(8.85xx10^(-6))=1.13xx10^(9) :. pH = - log [H^(+)]=-log (1.13xx10^(-9))=9-0.053=8.95` Alternatively, the given solution is a buffer solutionof a weak base and is salt with a strong acid (discussed in SEC. 7.31). For such a buffer, `pOH=pK_(b)+log.(["Salt"])/(["Base"])=4.75+log. (0.2)/(0.1) = 4.75+log 2 = 4.75 + 0.30 = 5.05 `[ Eqn . (vi)] `pH = 14 -pOH = 14 - 5.05=8.95` |
|
| 44. |
Calculate the pH of a solution containing 0.1 M acetic acid and 0.1 M benzoic acid. K_(a) for CH_(3)CO OH and C_(6)H_(5)COOH are 1.8xx10^(-5) and 6.5xx10^(-5) respectively . |
|
Answer» Solution :`[H_(3)O^(+)]=SQRT((1.8xx10^(-5))(0.1)+(6.5xx10^(-5))(0.1))` `=sqrt(1.8xx10^(-6)+6.5xx10^(-6))=sqrt(8.3xx10^(-6))=2.88xx10^(-3)M` PH `=-LOG [H_(3)O^(+)]=- log (2.88xx10^(-3))=3-0.4594=2.5406`. |
|
| 45. |
Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionization constants of acetic acid is 1.75xx10^(-5). Also calculate the change in pH of the buffer if to 1 litre of the buffer (i) 1 cc of 1 M NaOH are added (ii) 1 cc of 1 M HCl are added. Assume that the change in volume is negligible. What will be the buffer index of the above buffer ? |
|
Answer» Solution :`pH = pK_(a) + log. (["Salt"])/(["Acid"]) = - log (1.75 xx 10^(-5)) + log. (0.15)/(0.10)` `=(5-0.2430) + 0.1761=4.757+0.1761=4.933`. (i) 1 CC of 1 M NaOH contains NaOH `= 10^(-13)` MOL. This will convert `10^(-3)` mol of acetic acid into the salt so that salt formed `=10^(-3)` mol Now,[Acid]=0.10 - 0.001 = 0.099 M [Salt]=0.15+0.001 = 0.51 M `pH = 4.757 + log. (0.151)/(0.099) = 4.757 + 0.183 = 4.940` `:.` Increase in pH = 4.933 = 0.007 which is negligible. (ii) 1 cc of 1 M HCl contains HCl = `10^(-3)` mol. This will convert `10^(-3)` mol `CH_(3)CO ONa ` into `CH_(3)CO OH`. `:. ` Now, [Acid]=0.10 + 0.001 = 0.101 M [Salt] = 0.15 - 0.001 = 0.19 M `:. pH = 4.757 + log. (0.149)/(0.101) = 4.757+0.169= 4.926` `:. Decrease in pH = 4.933 - 4.926 = 0.007which is again negligible. (III) Calculation of buffer INDEX ltbr. No. of moles of HCl or NaOH added = 0.001 mol Change in pH = 0.007 Hence, buffer index `=(dn)/(dpH)=(0.001)/(0.007 ) = (1)/(7) = 0.143` |
|
| 46. |
Calculate the pH of a solution containing 2 g caustic soda/litre of water. |
|
Answer» `pH = - log (2xx10^(-13))` |
|
| 47. |
Calculate the pH of a buffer mixture of 0.05 M NH_(4)Cl and 0.12 M NH_(4)OH at 298 K. (Dissociation constant of ammonium hydroxide at 298 K is 1.8xx10^(-5)). |
|
Answer» Solution :`pK_(b)=-logK_(b)` = `-log1.8xx10^(-5)` = -(5.2553) = 4.7447 POH = `pK_(b)+"log"(["salt"])/(["BASE"])` = `4.7447+"log"(0.05)/(0.12)` = 4.74 - 0.38 = 4.36 `thereforepH=14-pOH` = 14 - 4.36 = 9.64 |
|
| 48. |
Calculate the pH of a 5.0 M H_(3)PO_(4) solution and the equilibrium concentrations of the species H_(3)PO_(4), H_(2)PO_(4)^(2-) and PO_(4)^(3-).(K_(a_(1))=7.5xx10^(-5)xx10^(-3), K_(a_(2))=6.2xx10^(-8), K_(a_(3))=4.2xx10^(-13)) |
|
Answer» Solution :`{:(,H_(3)PO_(4)+H_(2)O,hArr,H_(3)O^(+),+,H_(2)PO_(4)^(-)),("Initial conc.",5.0 M ,,,,),("Conc. after disso.",5-x,,x,,x),(,,,,,):}` `K_(a_(1))=([H_(3)O^(+)][H_(2)PO_(4)^(-)])/([H_(3)PO_(4)])=(x.x)/(5-x)=(x^(2))/(5)=7.5xx10^(-3)` or `x^(2)=37.5xx10^(-3)=3.75xx10^(-2)` or `x=1.94 xx 10^(-1) = 0.194` `[H_(3)O^(+)]=[H_(2)PO_(4)^(-)]=0.194M` `[H_(3)PO_(4)]=5-x=5-0.194=4.806M` As `K_(a_(2)) and K_(a_(3))` are `lt lt K_(a_(1))`, hence `[H_(3)O^(+)]` is mainly from this STEP. Hence, `pH = - log [H_(3)O^(+)]=- log 0.194 = 0.71` Step 2 of dissociation : `{:(,H_(2)PO_(4)^(-)+,H_(2)O,hArr,H_(3)O^(+)+,HPO_(4)^(2-)),("Initial",0.194M,,,,),("After dissociation",0.194-y,,,0.194+y,y),(,,,,~=0.194,):}` `K_(a_(2))=(0.194xxy)/(0.194-y)~=(0.194y)/(0.194)=y=6.2xx10^(-8)` Hence, `[HPO_(4)^(2-)]=6.2xx10^(-8)` Step 3 of dissociation : `{:(,HPO_(4)^(2-)+,H_(2)O,hArr,H_(3)O^(+)+,PO_(4)^(3-)),("Initial",6.2xx10^(-8)M,,,0.194+Z,z),("After disso.",6.2xx10^(-8)-z,,,~=0.194,),(,,,,,):}` `K_(a_(3))=(0.194xxz)/(6.2xx10^(-8)-z)~=(0.194z)/(6.2xx10^(-8))=4.2xx10^(-13)` or `z=(4.2xx10^(-13)xx6.2xx10^(-8))/(0.194)=134xx10^(-21)=1.34xx10^(-19)` `:. [PO_(4)^(3-)]=1.34xx10^(-19)` (negligible) Note that in the 2nd step of dissociation, concentration of anion is equal to `K_(a_(2))` value. |
|
| 49. |
Calculate the pH of a 0.10 M ammonia solution . Calculate the pH after 50.0 mL of this solutionis treated with 25.0 mL of 0.10 M HCl. The dissociation constant of ammonia, K_b=1.77xx10^(-5) |
|
Answer» Solution :Calculation of pH of 0.1 M ammonia : Ammonia is a WEAK base initially `[NH_3]`= 0.1 M In it X M is ionized So, At equilibrium . `[NH_3]=(0.1 -x) M=0.1` M `[NH_4^+]=[OH^-]`= x M `{:(,NH_(3(AQ)) + H_2O_((l)) hArr, NH_(4(aq))^(+)+, OH_((aq))^(-)),("At equilibrium :", (0.1-x), x M, x M),(, =0.1 M,,) :}` `K_b=([NH_4^+][OH^-])/0.1 = x^2/0.1` `therefore x=sqrt(0.1xxK_b)` `=sqrt(0.1xx1.77xx10^(-5))` `=sqrt(1.77xx10^(-6))` `=1.3304xx10^(-3) M =[OH^-]` pOH=-log `[OH^-]` =-log `(1.3304xx10^(-3))` =-(-2.876)=2.876 pH = 14.0-pOH=14.0-2.876 = 11.124 pH of mix solution after neutralization : Base = (50 ml 0.1 M `NH_3`) So mol of Base = `0.1 xx 50 / 1000 =5/1000` Acid =(25 mL 0.1 M HCl) So, mol of Acid = `0.1 xx 25/1000=2.5/1000` After Neutralization 25 mL `NH_3`= 75 mL In neutralization , Acid is consumed and base is remaining, (After Neutralisation) = (Initial molof Acid ) - (Initial mol of base ) `=(5/1000-2.5/1000)=2.5/1000` mol Molarity of remaining = `"mol"/"Total volume"` `=(2.5xx1000)/(1000xx75)`=0.0333 M Here, Total volume =(50+25) = 75 mL = 75/1000 L `NH_3 + H_2O hArr NH_4OH` `{:(,NH_(3(aq))+H_2O_((l)) hArr ,NH_(4(aq))^(+) +, OH_((aq))^(-)),("Initial:", 0.0333 M, 0.0333, 0.0),("Molarity at equilibrium",(0.0333-y),+y,yM),(,approx 0.0333 ,=0.0333 , ):}` `K_b=([NH_4^+][OH^-])/([NH_4OH])` `1.77xx10^(-5)=((0.033)[OH^-])/(0.033)` `therefore [OH^-]=1.77xx10^(-5)` Calculation of `[H^+]` : `[H^+]=K_w/([OH^-])=(1.0xx10^(-14))/(1.77xx10^(-5))` `=5.6497xx10^(-10)` Calculation of pH: `pH=-log [H^+]=-log (5.6497xx10^(-10))` =9.2480=9.25 |
|
| 50. |
Calculate the pHof a 0.01 Nsolution of acetic acid. K_(a) for CH_(3)CO OH is 1.8xx10^(-5) at 25^(@)C. |
|
Answer» Solution :`CH_(3)CO OH ` ionizes as : `CH_(3)CO OH + H_(2)O hArr CH_(3)CO O^(-) + H_(3)O^(+)` Applying the law of chemical 3quilibrium, we get `K_(a) = ([CH_(3)CO O^(-)][H_(3)O^(+)])/([CH_(3)C O OH ])` But `[CH_(3)CO O^(-)]=[H_(3)O^(+)]` [`:'` one molecule of `CH_(3)COOH` gives one `CH_(3)CO O^(-)` ion and one `H^(+)` ion or `H_(3)O^(+)` ion ] `:. K_(a) = ([H_(3)O^(+)]^(2))/([CH_(3)CO OH]) or [H_(3)O^(+)]=sqrt(K_(a)[CH_(3) CO OH])` But `K_(a) = 1.8xx10^(-5) and [CH_(3)CO OH]=0.01 M = 10^(-2)M` `:. [H_(3)O^(+)]=sqrt((1.8xx10^(-5))xx(10^(-2)))=sqrt(1.8xx10^(-7))=sqrt(18xx10^(-8))=sqrt(18) xx10^(-4) g ` ions/litre `= 4.242xx10^(-4)` gions/litre `:. pH = - LOG [H_(3)O^(+)]=-log(4.242xx10^(-4))=-(0.6276-4)=3.3727=3.37`. Alternatively, as DERIVED above, `[H_(3)O^(+)]={K_(a) [CH_(3)CO OH ]}^(1//2)=(K_(a)C)^(1//2)` `:. log [H_(3)O^(+)]=1/2(log K_(a) + log C ) or - log [H_(3)O^(+)]=1/2 (- log K_(a) - log C)` i.e., `pH=1/2(pK_(a) - log C)` This formula can be used directly for finding the pH of a weak acid solution of known concentration. Similarly, for a weak base, `pOH = 1/2 (pK_(b) - log C)` |
|