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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. |
The pH scale provides a convenient way to express the acidity and basicity of dilute aqueous solution. It depends upon the degree of dissociation of acids and bases in water Ionic product of water in the product of molar concentration of H^+ And OH^- ions at equilibrium at a given temperature. At 25^@C, K_w is 10 xx 10^(-14). The salts of weak bases and strong acids or weak acids and strong bases also give acidic or basic solution due to hydrolysis. Calculate the pH of a solution formed by mixing equal volumes of two solutions having pH =6 and pH=5 respectively. |
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Answer» SOLUTION :pH of solution A=6 `[H_3O^+]` of solution A = `10^(-6)` M pH of solution B =5 `[H_3O^+]` of solution B=`10^(-5)` M On mixing 1L of each solution , total VOLUME =2L `[H_3O^+]` in mixture =`(10^(-6)+10^(-5))/2` `=(10^(-5)(0.1+1))/2` `=(1.1xx10^(-5))/2 =5.5xx10^(-6)` pH =-LOG `(5.5xx10^(-6))=-0.740+6=5.26` |
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| 2. |
The pH scale provides a convenient way to express the acidity and basicity of dilute aqueous solution. It depends upon the degree of dissociation of acids and bases in water Ionic product of water in the product of molar concentration of H^+ And OH^- ions at equilibrium at a given temperature. At 25^@C, K_w is 10 xx 10^(-14). The salts of weak bases and strong acids or weak acids and strong bases also give acidic or basic solution due to hydrolysis. Will the pH of aqueous CuSO_4solution less than or more than or equal to 7? |
| Answer» SOLUTION :Aqueous `CuSO_4` solution is acidic and HENCE its PH < 7 . | |
| 3. |
The pH scale provides a convenient way to express the acidity and basicity of dilute aqueous solution. It depends upon the degree of dissociation of acids and bases in water Ionic product of water in the product of molar concentration of H^+ And OH^- ions at equilibrium at a given temperature. At 25^@C, K_w is 10 xx 10^(-14). The salts of weak bases and strong acids or weak acids and strong bases also give acidic or basic solution due to hydrolysis. 1 mL of 0.01 M HCl is added to 1L of sodium chloride solution. What will be the pH of the resulting solution? |
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Answer» Solution :NACL is neutral , it simply DILUTES the HCl solution from 1 mL to 100 mL so that `[H^+]=0.01/1000=10^(-5)` M pH =-log `(10^(-5))` =5. |
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| 4. |
The pH scale provides a convenient way to express the acidity and basicity of dilute aqueous solution. It depends upon the degree of dissociation of acids and bases in water Ionic product of water in the product of molar concentration of H^+ And OH^- ions at equilibrium at a given temperature. At 25^@C, K_w is 10 xx 10^(-14). The salts of weak bases and strong acids or weak acids and strong bases also give acidic or basic solution due to hydrolysis. What happens to ionic product of water at 27^@ C if some acid is added to it? |
| Answer» SOLUTION :IONIC PRODUCT will REMAIN UNCHANGED. | |
| 5. |
The pH of which compound in aqueous solution does not depend on its concentration in solution: |
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Answer» `underset( NH_2) underset(|) (RCH).COOH` |
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| 6. |
Total emf produd in a thermocouple does not depend on |
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Answer» `underset( NH_2) underset(|) (RCH).COOH` |
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| 7. |
The P^(H)of a weak mono basic acid is 5. The degree of ionisation of acid in 0.1 M solution is |
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Answer» `10^(-4)` |
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| 8. |
The pH of the solution produced by mixing equal volumes of 2.0xx10^(-3) MCHlO_(4) and 1.0xx10^(-2) MKClO_(4) is |
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Answer» `2.7` `[H^(+)]=(2xx10^(-3))/(2) :. pH = - log (10^(-3))=3.0`. |
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| 9. |
The pH of the solution formed on mixing 20 mL of0.05 MH_(2)SO_(4) with 5.0 mL of 0.45 M NaOH of 298 K is |
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Answer» 6 20 mLof 0.05 M `H_(2)SO_(4)=20xx0.05` millimoles = 1 millimole 5 mL of 0.45 M NaOH `= 5 xx 0.45` millimoles `=2.25` millimole 2 millimoles of NaOH REACT with 1 millimole of `H_(2)SO_(4)` `:.` NaOH left in the solution = 0.5 millimole Volume of solution = 25 mL `:. [OH^(-)]= (0.5)/(25)M = 0.01 M = 10^(-2)M` `:. [H^(+)]=10^(-12) ` or pH = 12 |
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| 10. |
The pH of the resultant solution of 20 ml of 0.1 M H_3PO_4 and 20 ml 0.1 M Na_3PO_4is |
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Answer» ` pK_(A1) +log 2 ` ` pH =P^(Ka_2) +log ""(2)/(2), pH =p^(Ka_2)` |
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| 11. |
The pH of the neutralisation point of 0.1 N NaOH with 0.1 N HCl is |
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Answer» 1 |
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| 12. |
The pH of the mixture of 25 ml of 0.01 M of CH_3 COOH (K_a of CH_3 COOH = 5 xx 10 ^(-5)) and 25 ml of 0.01 M of NaOH solution is |
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Answer» Solution :complete neutralization ` pH =7 +((pK_a+log c)/( 2)) ` ` = 7+ (( 5- log 5 + log ""(0.01)/(2))/( 2)) =7+((5-3)/(2))=8` |
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| 13. |
The pH of sulphuric acid solution is 2.3. What is the molar concentration of the acid ? |
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Answer» Solution :For sulphuric acid solution, pH = 2.3 `[H^(+)] = 10^(-pH) = 10^(-2.3)= 10^(-3)xx 10^(0.7)= 5 xx 10^(-3) mol L^(-1)` For sulphuric acid solution, `[H^+ ] = 2[H_2 SO_4 ]` MOLAR concentration of `H_2 SO_4 =([H^(+)])/(2)= ( 5 xx 10^(-3))/(2) = 2.5 xx 10^(-3) mol L^(-1)` |
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| 14. |
The P^(H)of solution is 9. It is times more basic than a solution with P^(H)= 6. |
| Answer» SOLUTION :1 unit gap = 10 TIMES , 3 units = 1000times | |
| 15. |
The pH of saturated solution of Ca(OH)_2 is 12.25. Then calculate its solubility product. |
| Answer» SOLUTION :`2.25xx10^(-5)` | |
| 16. |
The pH of rain water is ………....... |
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Answer» 6.5 |
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| 17. |
The pH of pure water at 25^(@)C and35^(@)C are 7 and 6 respectively. Calculate the heat of dissociation of H_(2)O into H^(+) and OH^(-) ions. |
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Answer» Solution :The dissociation reaction is : `H_(2)O hArr H^(+)+OH^(-)` At `25^(@)C`, pH = 7 means `[H^(+)]=10^(-7)M :. K_(w)=10^(-14)` At `35^(@)C`, pH =6 means `[H^(+)]=10^(-6)M :. K_(w)=10^(-12)` As equilibrium CONSTANT for the dissociation of `H_(2)O` are in the same ratio as ionic products of water, we can apply the relation `LOG.(K_(w_(2)))/(K_(w_(1)))=(Delta H)/(2.303 R)((1)/(T_(1))-(1)/(T_(2))):. log. (10^(-12))/(10^(-14))=(Delta H)/(2.303xx8.314 JK^(-1)"mol"^(-1))((1)/(298K)-(1)/(3-8K))` or `DeltaH=52898 J "mol"^(-1) = 52.898 kJ "mol"^(-1)`. |
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| 18. |
The P^(H) of pure water at 25^(0)C and 35^(0)C are 7 and 6 respectively. The heat of formation of water from H^(+) and OH^(-) will be. |
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Answer» `+84.551 KCAL "MOLE"^(-1)` At, `35^(0)C, [H^(+)] = 10^(-6)` `"log"(K_(omega_(2)))/(K_(omega_(1))) = (DeltaH)/(R)[(T_(2) - T_(1))/(T_(1) xx T_(2))]` |
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| 19. |
The pH of normal rain water is |
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Answer» 6.5 |
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| 20. |
The pH of neutral water at 25^(@)C is 7.0 . As the temperature increases, ionisation of water increases, however, the concentration of H^(+) ions and OH^(-) ions are equal. What will be the pH of pure water at 60^(@)C ? |
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Answer» Equal to 7.0 |
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| 21. |
The pH of neutral water at 25^@C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H ions and OH^-ions are equal. What will be the pH of pure water at 60^@C ? |
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Answer» Equal to 7.0 As `25^@C[H^+]=[OH^-]=10^(-7)` `K_w=[H^+][OH^-]=10^(-14)` On heating , `K_w` increases, i.e., `[H^+][OH^-] gt 10^(-14)` As `[H^+]=[OH^-]` or , `[H^+]^2 gt = 10^(-14)` or `[H^+] gt 10^(-7)` M `therefore pH lt 7` With rise in temperature, pH of pure water decreases and it become less than 7 at `60^@`C. |
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| 22. |
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each. |
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Answer» Solution :(i)Calculation of `[H^+]` in milk [pH=6.8] pH=-log `[H^+]`=6.8 = `bar7.2` `therefore [H^+]` = Antilog `(bar7.2)=1.5849xx10^(-7)` `approx 1.58xx10^(-7)` M (II)`[H^+]` CONCENTRATION in BLACK coffee [pH=0.5] pH=-log `[H^+]`-5.0 and log `[H^+]`=-5 So, `[H^+]` = Antilog `[H^+]` =Antilog (-5) `=1xx10^(-5)` M (iii)Concentration of `[H^+]` in tomato juice [pH=4.2] pH=-log `[H^+]` =4.2 log `[H^+]`=-4.2 = `bar5.8` `therefore [H^+]` Antilog `(bar5.8)=6.3096xx10^(-5)` `approx 6.31xx10^(-5)` M (iv)concentration of `[H^+]` in lemon juice [pH=2.2] pH=-log `[H^+]` = 2.2 `therefore` log `[H^+]` = 2.2 =`bar3.8` `therefore [H^+]` =Antilog `(bar3.8)=6.3096xx10^(-3)` `approx 6.31xx10^(-3)` M (v) Concentration of `[H^+]` in egg white [pH=7.8] pH =-log `[H^+]`=7.8 `therefore` log `[H^+]`=-7.8 =`bar8.2` `therefore [H^+]`=Antilog `(bar8.2)=1.5849xx10^(-8)` `approx 1.58xx10^(-8)` M |
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| 23. |
The pH of HCl solution is 2. When it is diluted with equal volume of water, what is the pH of dilute acid solution ? |
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Answer» Solution :Given PH of HCL solution = 2. PROTON concentration of the solution `= 10^(-pH) =10^(-2) M` When diluted with equal VOLUME of water, the proton concentration `[H^(+)]=(Vxx 10^(-2) )/(V+V) = 5 xx 10^(-3) ।pH=- LOG(5 xx 10 ^(-3)=3- log=2.3.` |
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| 24. |
The P^(H) of HCI is 5. It is diluted by 1000 times. Its P^(H) will be |
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Answer» `5` DILUTED by 1000 times , So PH =8 ,But acids cannot have `pH =8 , THEREFORE pH =6-7` |
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| 25. |
The P^(H)of HCI is 3. Then the P^(H) of NaOH solution having same molar concentration is |
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Answer» Solution :` pH =3 , [H^(+) ] =10 ^(-3)N` `[ OH^(-) ] = 10 ^(-3)N , pOH =3 RARR pH = 11` |
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| 26. |
The P^(H) of HCI is 1. The amount of NaOH to be added to 100 ml of such a HCI solution to get p^(H) of 7 is |
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Answer» 4g no. of EQ. . Of BASE = no. of e eq. of acid pH of HCl = 1 ` [H^(+)= 0.01N , [HCl]= 0.1 N` From(1) , [NaOH] = 0.1 N ` 0.1 = (W)/( 40 ) xx ( 1000)/(100)THEREFORE W = 0.4 gm ` |
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| 27. |
The pH of buffer solution remain same when any amount of dilution is done. |
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Answer» |
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| 28. |
The ph of blood stream is maintained by a proper balance of H_(2)CO_(3) and NaHCO_(3) concentrations. What volume of 5M NaHCO_(3) solution, should be mixed with 10mL sample of blood which is 2M in H_(2)CO_(3) in order to maintain a pH of 7.4.K_(a) for H_(2)CO_(3) in blood is 4.0 xx 10^(-7)? |
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Answer» Solution :`[H_(2)CO_(3)]` in blood `=2M [NaHCO_(3)] = 5M` VOLUME of blood `= 10mL` Let volume of `NaHCO_(3)` used `= VmL` `[H_(2)CO_(3)]` in mixture =` (2xx10)/((V +10))` `[NaHCO_(3)]` in mixture `= ((5xxV))/((V +10))` `pH = pK_(a) + log .(["salt"])/(["acid"])` `7.4 = - log 4.0 XX 10^(-7) + "log"((5xxV)//(V+10))/((2xx10)//(V+10))` `V = 40mL` |
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| 29. |
The pH of blood stream is maintained by aproper balance of H_(2)CO_(3) and NaHCO_(3) concentrations. What volume of 5 M NaHCO_(3) solution should be mixed with a 10 ml sample of blood which is 2 M in H_(2)CO_(3) in order to maintain a pH of 7.4 ? K_(a) for H_(2)CO_(3) in blood is 7.8xx10^(-7). |
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Answer» Solution :`pH = pK_(a) + LOG. (["Salt"])/(["Acid"])` `7.4= - log. (7.8xx10^(-7)) + log. ([NaHCO_(3)])/([H_(2)CO_(3)])` `log .([NaHCO_(3)])/([H_(2)CO_(3)])=7.4+ (-7+0.8921)=1.2921 or ([NaHCO_(3)])/([H_(2)CO_(3)]) = ` Antilog1.2921 = 19.59 `H_(2)CO_(3)` present in 10 ML of blood `(2 M H_(2)CO_(3))=(2)/(1000)xx10=0.02` mole `:. NaHCO_(3)` that should be present `=19.59xx0.02 = 0.3918 ` mole Volume of 5 M `NaHCO_(3)` required `= (1000)/(5) XX 0.3918 = 78.36` ml |
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| 30. |
The pH of blood serum is 7.4 . What is the hydrogen ion concentration of blood serum ? |
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Answer» |
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| 32. |
The pH of an enzyme catalysed reaction has to be monitored between pH 7-8. What indicator should be used to monitor and control the pH of the reaction ? |
| Answer» SOLUTION :BROMOTHYMOL BLUE (6.0 – 7.5). | |
| 33. |
The pH of aqueous NaOH is 13. What weight of solute is present in 500 ml solution ? |
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Answer» Solution :GIVEN PH = 13, pOH = 14 - 13 = 1, `[OH^(-)] = [NaOH] = 10^(-1)=0.1 M` Gram molecular mass (GMW) of NaOH = `40 g mol^(-1)` Weight of solute present in IL solution=`M xxGMV =(1)/( 10 ) xx 40 =4` The weight of NaOH present in 500 ml solution = 2 grams. |
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| 34. |
The pH of an enzyme catalysed reaction has to be maintained between 7 and 8. What indicator should be used to monitor and a controlthe pH ? |
| Answer» Solution :BROMOTHYMOL BLUE or PHENOL RED or Cresol red . | |
| 35. |
The P^(H)of an aqueous solution of NH_4 CN( K_a of HCN is 9.2 xx 10^(-10)& K_b of NH_4 OHis 1.8 xx 10^(-5)) |
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Answer» Solution :` pH =7+ ( (PKA- pKb)/(2) )` ` =7+[( (10 -LOG 9.2)- (5 - log 1.8))/(2) ]` |
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| 36. |
The pH of an aqueous solution is 6.58. Calculate the concentration of hydroxyl ion of the given solution. |
| Answer» SOLUTION :`3.8 XX 10^(-8) M` | |
| 37. |
The pH of acid rain may be |
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Answer» 8.2 |
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| 38. |
the pH of a solution prepared by mixing 2M, 100 mL HCl and M, 200 mL NaOH at 25^(@)C is |
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Answer» 8 |
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| 39. |
The pH of a solution of H_2O_2is 6.0. Some chlorine gas is bubbled into this solution. Which of the following is correct? |
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Answer» The pH of resultant solution BECOMES 8.0 |
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| 40. |
The pH of a solution obtained by mixing 50 mL of 0.2 MHCl with 50 mL of 0.20 M CH_(3) C O OH is |
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Answer» `0.30` 50 ML of 0.2 M HCl=`50xx0.2` millimoles = 10 millimoles Volume ofsolution after mixing = 100 mL `:.` MOLAR conc.of HCl `=(10)/(100)=0.1M = 10^(-1)M` `:. pH=1` |
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| 41. |
The pH of a solution obtained by mixing 100 mL of a solution pH=3 with 400 mL of asolution of pH=4 is |
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Answer» 3- log 2.8 `10^(-3)xx100+10^(-4)xx400=N_(3)(100+400)` or, `N_(3)=(0.1+0.04)/(500)=(0.14)/(500)=2.8xx10^(-4)M` `pH =-log (2.8xx10^(-4))=4-"log"2.8` |
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| 42. |
The pH of a solution obtained by dissolving 0.1 mole of an acid HA is 100 ml oftheaqueous solution was found to be 3.0 . Calculate the dissociation constant of the acid. |
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Answer» Solution :`PH = - log [H_(3)O^(+)] :. Log [H_(3)O^(+)]=-pH = - 3.0 ` [pH = 3.0 , GIVEN] or `[H_(3)O^(+)]` = antilog (-3) = antilog `bar(3) = 10^(-3)` g ions/litre = 0.001 g ion/litre Original CONC.of the acid HA = 0.1 mole in 100 ml. = 1 mole/litre `{:("HA dissociates as :",HA""+""H_(2)O,hArr,H_(3)O^(+)+,A^(-)),("Initial conc. :",1 M,,0,0),("Conc. at equilibrium:",1-0.001M,,0.001M,0.001M),(,,,,):}` `:.` DISSOCIATION constant (K) will be given by `K=([H_(3)O^(+)][A^(-)])/([HA])=(0.001xx0.001)/(1-0.001)=(10^(-6))/(1)=10^(-6)` (Neglecting 0.001 in comparison to 1) |
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| 43. |
The P^(H )of a solution is 6. Its [H_(3),O^(4)] is decreased by 1000 times. Its P^(H) will be |
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Answer» `9` |
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| 44. |
The pH of a solution is 5.0 . To a 10 ml of this solution , 990 ml of water is added . Then the pH of the resulting solution is |
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Answer» 7 |
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| 45. |
The pH of a solution is 5. Its hydrogen ion concentration is increased 100 times. What is the pH of the resulting solution ? |
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Answer» |
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| 46. |
The pH of a solution is 4.301 . To 50 ml of this solution 200 ml of water is added . Then the pH of resultingsolution is |
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Answer» 4.9 |
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| 47. |
The P^(H)of a solution is 3.0. This solution is diluted by 100 times. Then the P^(H)of the resulting solution is |
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Answer» 5 ` C = [H^(+) ] =10 ^(-3)N rArr N_1V_1 =N_2V_2` ` 10 ^(-3) XX V = N_2 xx 100 V ` `N_2=10 ^(-5)= [H^(+) ] rArr pH = 5` |
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| 48. |
The P^(H) of a solution is 3.602. Its H^(+) ion concentration is |
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Answer» ` 4 XX 10^(-14)` Anti log ` (0. 398 ) xx 10 ^(-4)= 2.5 xx 10^(-4)` |
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| 49. |
The pH of a solution is 12 . The no. of H^(+) ions present in 1 ml of this solution at 25^(@) C is |
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Answer» `6.02 xx 10^(20)` |
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