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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 401. |
A saturated solution of silver benzoate, `AgOCOC_(6)H_(5)` has a `pH` of `8.63, K_(a)` for benzoic acid is `6.5 xx 10^(-5)`. Estimate `K_(sp)` for silver benzoate. |
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Answer» `pH = 8.63, pOH = 5.37, [overset(Θ)OH] = 4.3 xx 10^(-6)` `{:(,C_(6)H_(5)COO^(Θ)+H_(2)Orarr,C_(6)H_(5)COOH+,overset(Θ)OH,),("Initially",C,0,0,),("Final",C(1-h),Ch,Ch,):}` `K_(h) = ([C_(6)H_(5)COOH][overset(Θ)OH])/([C_(6)H_(5)COO^(Θ)] ),Ch = 4.3 xx 10^(-6)` `:. [C_(^)H_(5)COO^(Θ)] = ((4.3xx10^(-6))(4.3xx10^(-6)))/(K_(h)) ...(i)` `K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(6.5 xx 10^(-5))` Substituting the value of `K_(h)` in equaiton (i) `:. [C_(6)H_(5)COO^(Θ)] = ((4.3xx10^(-6))^(2)xx(6.5xx10^(-5)))/(10^(-14))` `C_(6)H_(5)COOAg hArr C_(6)H_(5)COo^(Θ) + Ag^(o+)` `K_(sp) = [Ag^(o+)] [C_(6)H_(5)COO^(Θ)] = (0.12)^(2) = 1.4 xx 10^(-2)` Second method: Use formula for the `pH` of salt of `S_(B)//W_(A))` `pH = (1)/(2) (pK_(w) + pK_(a) + logC)` `8.63 = (1)/(2) (14 + 4.187 + logC)` `:. C = 0.12` `K_(sp) = [Ag^(o+)] [C_(6)H_(5)COO^(Θ)] = (0.12)^(2) = 1.4 xx 10^(-2)` |
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| 402. |
100 mL solution (I) of buffer containin 0.1 (M) HA and 0.2 (M)`A^-`, is mixed with another solution (II) of 100 mL containing 0.2 (M) HA and 0.3 (M) `A^-` . After mixing, what is the pH of resulting solution ? [Given: `pK_a` of HA=5]A. 5-log`5/3`B. 5+log`5/3`C. 5+log`2/5`D. 5-log`5/2` |
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Answer» Correct Answer - b |
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| 403. |
How many mole of HCI will be required to prepare one litre of buffer solution (containing `NaCN+ HCI`) of `pH 8.5` using `0.01`g formula weight of NaCN ? `K_(HCN)=4.1xx10^(-10))` |
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Answer» `NaCN+HCI` is not a buffer but if HCI is in less amount then it gives a buffer as it produces HCN. `{:(,NaCN+,HCIrarr,NaCI+,HCN),("Mole added", 0.01,alpha,0,0),("Mole after reaction", (0.01-alpha),0,alpha,alpha):}` This is buffer of HCN + NaCN. Let a mole of HCI be used for this purpose. `:. pH= -log K_(a)+log((0.01-a)/(a))` `8.5= -log 4.1xx10^(-10)+log ((0.01-a)/(a))` `:. a = 8.85xx10^(-3)` mole of HCI` |
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| 404. |
What is the pH of a solution that is 0.20 M in HF and 0.40 M in NaF ? `[K_a=7.2xx10^(-4)]`A. 1.92B. 2.84C. 3.14D. 3.44 |
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Answer» Correct Answer - d |
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| 405. |
In what volume ratio should you mix `1.00 M` solution `NH_(4)Cl` and 1.00 M `NH_(3)` to produce a buffer solution of pH 9.80 ? `[pK_(b)(NH_(3))=4.74]`A. `1:1.35`B. `3.5 :1`C. `2:1`D. `1:2` |
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Answer» Correct Answer - B `[NH_(4)Cl]=V_(1)mL =V_(1)` millimol `[NH_(3)]=V_(2)mL =V_(2)` millimol `=pH =9.80 , pOH =4.2` `pOH =pK_(a)+log.(V_(1))/(V_(2))` `4.2 =4.74 +log. (V_(1))/(V_(2))` `-0.54=log. (V_(1))/(V_(2))` or `log.(V_(2))/(V_(1))=0.54` `(V_(2))/(V_(1))=` antilog `0.54=3.5` Thusl, `(V_(2))/(V_(1))=(3.5 )/(1)` |
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| 406. |
What volumes of 0.200 M `HNO_2` and 0.200 M `NaNO_2` are required to make 500 mL of a buffer solution with pH =3.00 ? [`K_a` for `HNO_2=4.00xx10^(-4)`]A. 250 mL of eachB. 143 mL of `HNO_2` and 357 mL of `NaNO_2`C. 200 mL of `HNO_2` and 300 mL of `NaNO_2`D. 357 mL of `HNO_2` and 143 mL of `NaNO_2` |
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Answer» Correct Answer - d |
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| 407. |
What is the pH of 1.00 L sample of a buffer solution containing 0.10 mol of benzoic acid and 0.10 mol of sodium benzoate of which 0.010 mol of NaOH has been added ? [`K_a` benzoic acid=`6.5xx10^(-5)`]A. 4.27B. 4.23C. 4.15D. `4.10` |
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Answer» Correct Answer - a |
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| 408. |
A student is asked to prepare a buffer solution with a pH of 4.00. This can be accomplished by using a solution containing which of the following ? A. `HNO_2` onlyB. HCN onlyC. `HNO_2` and `NaNO_2`D. HCN and NaCN |
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Answer» Correct Answer - c |
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| 409. |
A buffer solution made with `NH_3` and `NH_4Cl` has a pH of 10.0 which procedure could be used to lower the pH ? 1.Adding HCl , 2.Adding `NH_3` , 3. Adding `NH_4Cl`A. 1 onlyB. 2 onlyC. 1 and 3 onlyD. 2 and 3 only |
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Answer» Correct Answer - c |
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| 410. |
Which base is most suitable to prepare a buffer solution with a pH =11.00 ?A. Ammonia `(K_b=1.8xx10^(-5))`B. Aniline `(K_b=4.0xx10^(-10))`C. Methylamine `(K_b=4.4xx10^(-4))`D. Pyridine `(K_b=1.7xx10^(-9))` |
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Answer» Correct Answer - c |
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| 411. |
Which pair of solutes could be used to prepare an aqueous buffer solution with a pH `lt 7`?A. `HCI - NH_(4)CI`B. `HF-NaF`C. `NH_(3)-NH_(4)CI`D. `NaOH-NaCI` |
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Answer» Correct Answer - b |
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| 412. |
Which of the following solutions have `pH lt 7`.A. `BaI_(2)`B. `AI(NO_(3))_(3)`C. `CH_(3)COONH_(4)`D. `CsI` |
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Answer» Correct Answer - B Salt of `W_(B)//S_(A)` have `pH lt 7`. (Salt of `W_(A)//W_(B)` have `pH lt 7`, if acidic part is stronger, i.e., if `K_(a) gt K_(b)` or `pK_(b) gt pK_(a))` a. Salt of `Ba(OH)_(2)(S_(B)) HI (S_(A))` `:. pH = 7` b. Salt of `A1(OH)_(3) (S_(B))` and `HNO_(3)(S_(A))` ` :. pH lt 7`. c. Salt of `CH_(3)COOH (W_(A))` and `NH_(4)OH(W_(B))` (but `K_(a) = K_(b)`, so `pH = 7`. d. Salt of `Cs(OH)(S_(B))` and `HI (S_(A))` so `pH = 7`. |
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| 413. |
The `K_(a)` for formic acid and acetic acid are `2.1 xx 10^(-4)` and `1.1 xx 10^(-5)`, respectively. Calculate relative strength of acids |
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Answer» Relative strength of weak acids `= sqrt((K_(a_(1)))/(K_(a_(2)))) xx (C_(1))/(C_(2))` Assume `C_(1)` and `C_(2)` are same (although not given). `:.` Relative strength `= sqrt((K_(a_(1)))/(K_(a_(2)))) = sqrt((2.1 xx 10^(-4))/(1.1xx10^(-5)) = 4.36:1` Relative strength for `HCOOH` to `CH_(3)COOH = 4.36:1` |
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| 414. |
The `K_(a)` value of formic acid and acetic acid are respectively `1.77 xx 10^(-4)` and `1.75 xx 10^(-5)` . The ratio of the acid strength of 0.1 N acids isA. 100B. 3.178C. 0.3D. 0.1 |
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Answer» Correct Answer - B Relative strength of two weak monobasic acids of equal concentration is `("Strength of HCOOH")/("Strength of "CH_(3)COOH)=sqrt((K_(HCOOH))/(K_(CH_(3)COOH)))` `=sqrt((1.77 xx10^(-4))/(1.75 xx 10^(5)))` `=sqrt((17.7)/(1.75))=sqrt(10.11)=3.178` |
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| 415. |
`K_(a)` for formic acid and acetic acid are `2.1xx10^(-4)` and `1.1xx10^(-5)` respectively. The relative strength of acids is:A. `19:1`B. `2.3:1`C. `1:2.1`D. `4.37:1` |
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Answer» Correct Answer - d |
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| 416. |
`K_(a)` for formic acid and acetic acid are `2.1xx10^(-4)` and `1.1xx10^(-5)` respectively. The relative strength of acids is:A. `10:1`B. `1:10`C. `1:sqrt(10)`D. `sqrt(10):1` |
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Answer» Relative strengths of weak acids=`sqrt(((K_(a_(1)))/(K_(a_(2)))))` Assume `C_(1)` and `C_(2)` are same (Although not given). Relative strength = `sqrt(((K_(a_(1)))/(K_(a_(2)))))=sqrt(((1.8xx10^(-4))/(1.8xx10^(-5))))` Relative strength for `HCOOH` to `CH_(3)COOH=sqrt(10):1` |
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| 417. |
The ionisation constant of `HF` is `3.2 xx 10^(-4)`. a. Calculate the dergee of dissociation of aall species present `M` solution. b. Calculate the concentration of all species present `(H_(3)O^(o+),F^(Θ)` and `HF`) in the solution. c. Calculate method: |
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Answer» a. The following proton transfter reactions are possible, `HF +H_(2)O hArr H_(3)O^(oplus)+F^(Theta)(K_(a)=3.2 xx10^(-4))` ....(i) `H_(2)O+H_(2)O hArr H_(3)O^(oplus)+overset(Theta)(O)H (Kw=10^(14))`....(ii) Since `K_(a)gt gt K_(w)`, reaction (i) is the main reaction. `{:(,HF+, H_(2)O hArr,H_(3)O^(oplus),F^(Theta)),("Initial conc",rArr0.02,-,0,0),("Change in conc",rArr-0.02alpha,-,+0.02alpha,+0.02alpha),("Eq conc",rArr0.02-0.02alpha,-,,),(,=0.021(1-alpha),,0.02alpha,0.02alpha):}` `:.K_(a) = ((0.02alpha)^(2))/(0.02(1-alpha)) = (0.02alpha^(2))/(1-alpha) = 3.2 xx 10^(-4) ..(i)` Solving the equaiton (i), the following quadratic equaiton is obained, `alpha^(2) +1.6 xx 10^(-2)alpha - 1.6 xx 10^(-2) = 0`. `:. alpha = +0.12` and `-0.12`. (-ve value is not taken) Thus, `alpha = 0.12` b. `[H_(3)O^(o+)] = [F^(Θ)] = C alpha = 0.02 xx 0.12 = 2.4 xx 10^(-3) M` `[HF] = C(1-alpha) = 0.02 (1-0.12) = 17.6 xx 10^(-3)M` c. `pH = - log [H^(o+)] =- log (2.4 xx 10^(-3)) = 2.62` Note: `alpha - 0.12` which is greater than `0.05, (alpha ge 0.05` or `alpha le 5%)` So the term `(1-alpha) != 1`, is taken. Second method: By using formula of `alpha`, and `ph`. a. `alpha = sqrt((K_(a))/(C)) = sqrt((3.2 xx 10^(-4))/(0.02)) = sqrt(16xx10^(-4)xx10)` `= 4 xx 10^(-2) xx sqrt(10)` `(sqrt(10) = 3.162)` `= 4xx3.162 xx 10^(-2)` `= 0.126` `:. alpha= 0.12` b. `[H_(3)O^(o+)] = [F^(Θ)] = Calpha = 0.02 xx 0.12 = 2.4 xx 10^(-3)M` `[HF] = C(1-alpha) = 0.02 (1-0.12) = 1.76 xx 10^(-3)M` c. `pK_(a) =- log (3.2 xx 10^(-4)) =- log (32 xx 10^(-5))` `=- log (25 xx 10^(-5))` `=- 5 log 2+5` `=- 5 xx 0.3 +5 = 3.5` `pK_(a) ~~ 3.5` `pH = (1)/(2) (pK_(a) - logC) = (1)/(2) (3.5-log0.02)` `= (1)/(3) (3.5 - log 2 xx 10^(-2))` `= (1)/(2) (3.5 - 0.3 +2)` `= (1)/(2) (5.2) = 2.6` `:. pH ~~ 2.6`. |
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| 418. |
The vapour pressur of `0.01molal` solution of weak base `BOH` in water at `20^(@)C` is `17.536mm`. Calculate `K_(b)` for base. Aqueous tension at `20^(@)C` is `17.540mm`. Assume molatilly and molarity same. |
| Answer» Correct Answer - `9.74xx10^(-4);` | |
| 419. |
The solubility of AgCI in conc. HCI is……… than in water.A. moreB. lessC. sameD. either of these |
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Answer» Correct Answer - A Due to the formation of complex `AgCI+HCIrarr HAgCI_(2)` |
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| 420. |
`BOH` is a weak base, molar concentration of `BOH` that provides a `[OH]^(-)` of `1.5xx10^(-3) M[K_(b)(BOH)=1.5xx10^(-5) M]` isA. `1.5xx10^(-5)M`B. `0.015M`C. `0.0015`D. `0.15M` |
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Answer» Correct Answer - D `BOH hArr B^(+) + OH^(-) ` `K_(b)=([B^(+)][OH^(-)])/([BOH])` `[B^(+)]=[OH^(-)]` `:. K_(b)=([OH^(-)]^(2))/([BOH])` `[BOH]=([OH^(-)]^(2))/(K_(b))` `=((1.5 xx 10^(-3))^(2))/(1.5 xx 10^(-5))=1.5 xx 10^(-1)M` `=0.15 M` |
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| 421. |
Calculate for `0.01N` solution of sodium acetate, a. Hydrolysis constant b. Dergee of hydrolysis c. `pH` Given `K_(a) = 1.9 xx 10^(-5)` |
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Answer» For `{:(,CH_(3)COONa+,H_(2)OhArr,CH_(3)COOH+,NaOH),("Before1hysrolysis",1,,0,0),("After hydrolysis",1-h,,h,h):}` a. `:. K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(1.9xx10^(-5)) = 5.26 xx 10^(-10)` b. `h = sqrt((K_(h))/(c)) = sqrt((5.26xx10^(-10))/(0.01)) = 2.29 xx 10^(-4)` c. `[overset(Θ)OH]` from `NaOH`, a strong alkali `= Ch` `:. =0.01 xx 2.29 xx 10^(-4) = 2.29 xx 10^(-6)M` `:. pOH = 5.64` and `pH = 8.36` |
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| 422. |
Calculate pH for: (a) `0.01N Ca(OH)_(2)` (b) `0.01M Ca(OH)_(2)` (c ) `0.0008M Mg(OH)_(2)` Assume complete ionisation of each. |
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Answer» (a) `0.01N Ca(OH)_(2):` `{:(Ca(OH)_(2),rarrCa^(2+)+,2OH^(-)),(10^(-2)N,0,0),(0,10^(-2),10^(2-)):}` Eq. ratio of `Ca(OH)_(2): Ca^(2+):OH^(-1)::1:1:1` `(.: "Equivalent litre"^(-1)"are given")` `:. [OH^(-)]=10^(-2)M` `:. pOH=2` `:. pH=12` (b) `0.01M Ca(OH)_(2)`: `{:(Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),(10^(-2)M,0,0),(0,10^(-2),2xx10^(-2)):}` Mole ratio of `Ca(OH)_(2):Ca^(2+):OH^(-1):1:1:2` `:. [OH^(-)]=2xx10^(-2)M` `:. pOH=1.6989` `:. pH=14-1.6989=12.3011` (c ) `0.0008M Mg(OH)_(2):` `{:(Mg(OH)_(2)rarr,Mg^(2+)+,2OH^(-)),(8xx10^(-4)M,0,0),(0,8xx10^(-4),2xx8xx10^(-4)):}` `:.[OH^(-)]=16xx10^(-4)M` `:. pOH=2.7958` `:. pH=11.2042` |
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| 423. |
Statement: The pH of an aqueous solution of acetic acid remains unchanged on the addition of sodium acetate. Explanation: The ionisation of acetic acid is suppressed by the addition of sodium acetate.A. S is correct but E is wrong.B. S is wrong but E is correc.C. Both S and E are correct and E is correct explanation of S.D. Both S and E are correct but E is correct explanation of S. |
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Answer» Correct Answer - C Explanation is correct reason for stetement. |
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| 424. |
Calculate the pH of a solution made by mixing 50 mL of `0.01 Mba(OH)_(2)` with 50 mL water. (Assume complete ionisation) |
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Answer» Milli-mole of `Ba(OH)_(2)=50xx0.01=0.5`. The solution is diluted with `50 mL H_(2)O` and thus volume becomes 100 mL. Thus `[Ba(OH)_(2)]=(0.5)/(100)=0.005M` `{:(,Ba(OH)_(2)rarr,Ba^(2+)+,20H^(-)),("Initial conc", 0.005,0,0),("Final conc".,0,0.005,2xx0.005=0.01):}` `[OH^(-)]=0.01=1xx10^(-2)` `:. pOH=2` `:. pH=14-2=12` |
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| 425. |
How much the concentration of `Ag^(o+)`ions in a saturted solution of `AgCI` diminish if such an amount of `HCI` is added to it that the concentration of `CI^(Θ)` ions in the solution becomes equal to `0.03M`? Also find the amount of `AgCI` precipitated at the given concentration. `K_(sp)` of `AgCI = 1.8 xx 10^(-10)`. |
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Answer» `HCI` is added to a solution containing `Ag^(o+)` ions in saturated solution. First find the concentration of `Ag^(o+)` ion in this solution `AgCl(s) hArr Ag^(oplus) + Cl^(Theta)` `rArr K_(sp) [Ag^(o+)] [CI^(Θ)] = x^(2)` where `x` is solubility of `AgCI` in mol `L^(-1)` `rArr [Ag^(o+)] = sqrt(K_(sp)) = sqrt(1.8 xx 10^(-10)) = 1.34 xx 10^(-5)M` When `HCI` is added, the ionic product of `AgCI` approaches the `K_(sp)` value of `AgCI`, the precipiation of `Ag^(o+)` ions will occur. As ionis product increase (i.e., become greater than `K_(sp)` value), and appreciable amount of `AgCO` precipitates out, and precipitation continues till ionic product equals solubility product `(K_(sp))`. Ionic product `= [Ag^(o+)] [CI^(Θ)] = K_(sp)` `rArr [Ag^(o+)] = (K_(sp))/([CI^(Θ)]) = (1.8 xx 10^(-10))/((0.03)) = 6.0 xx 10^(-9)M` Now this is the amount of `Ag^(o)` ions left un-precipitated `rArr [Ag^(o+)]` diminishes in the soolution by `(6.0 xx 10^(-9))/(1.34 xx 10^(-5)) = ((1)/(2233))` times The concentration of `AgCI` precipitated out of the solution `= [Ag^(o+)]_("initial") - [Ag^(o+)]_("left")` `= 1.34 xx 10^(-5) - 6.0 xx 10^(-9)M` It means almost whole of `AgCI` is precipitated out of the solution at `[CI^(Θ)] = 0.03M`. |
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| 426. |
Calculate pH for: (a) `0.001 NaOH`, (b) `0.01N Ca(OH)_(2)`, (c ) `0.01M Ca(OH)_(2)`, (d) `10^(-8)M NaOH`, (e ) `10^(2)MNaOH`, (f) `0.0008MMg(OH)_(2)` Assume complete ionisation of each. |
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Answer» (a) `0.001N NaOH:` `{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(-3)N,0,0),(0,10^(-3),10^(-3)):}` `:. [OH^(-)]=10^(-3)M` `:. pOH= -log[OH^(-)]` `= -log 10^(-3)=3` `:. pH=14-pOH=14-3=11` `:. pH =11` (b) `0.01N Ca(OH)_(2)` `{:(Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),(10^(-2)N,0,0),(0,10^(-2),10^(-2)):}` `:. [OH^(-)]=10^(-2)M` `:. pOH=2` `:. pH=12` (c ) `0.01M Ca(OH)_(2)`: `{:(Ca(OH)_(2)rarr,Ca^(2+)+,2OH^(-)),(10^(-2)M,0,0),(0,10^(-2),2xx10^(-2)):}` `[OH^(-)]=2xx10^(-2)M` `:. pOH=1.6989` `:. pH=14-1.6989=12.3010` (d) `10^(-8)MNaOH:` `{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(-8)M,0,0),(0,10^(-8),10^(-8)):}` `:. [OH^(-)]=10^(-8)M` Now proceed for `OH^(-)` as in problem 19 part (e ). (e ) `10^(2)MNaOH:` `{:(NaOHrarr,Na^(+)+,OH^(-)),(10^(2),0,0),(0,10^(2),10^(2)):}` `[OH^(-)]=10^(2)M` Now proceed in problem 19 part (f). (f) `0.0008MMg(OH)_(2)`: `{:(Mg(OH)_(2)rarr,Mg^(2+)+,2OH^(-)),(8xx10^(-4)M,0,0),(0,8xx10^(-4),2xx8xx10^(-4)):}` `:. [OH^(-)]=16xx10^(-4)M` `:. pOH=2.7958` `pH=11.2041` |
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| 427. |
What will be the `pH` of `5xx10^(-6)MBa(OH_(2))` solution of `25^(@)C`? |
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Answer» `[OH^(-)]_("from strong base")=2(5xx10^(-6))=10^(-5)M` `,.pH=14-p(OH)=14-(-log[OH^(+)])=14-(-log10^(-5))` `14-5=9` |
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| 428. |
Calculate the `pH` of `500mL` Solution of `1 M BOH (K_(b)=2.5xx10^(-5))` |
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Answer» `{:(,BOH,hArr,B^(+),+,OH^(-)),(t=0,1M,,0,,0),(t=eq,[1-alpha],,alpha,,alpha):}` `:.(alpha^(2))/(1-alpha)=K_(b)` expecting `a ltlt1 :.alpha^(2)=2.5xx10^(-5)rArr alpha=5xx10^(-3)` `:.[OH^(-)]=5xx10^(-3)MrArr pOH=3-log5rArr pOH=2.3rArr pH=11.7` Note: (i) Volume of the solution given in questionhas no significance. (ii) After getting `alpha` value as negligible ,direct relation :`pOH=(1)/(2)(pK_(b)-logC)` could also be used . |
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| 429. |
`500mL` of `10^(-5)MNaOH` is mixed with `500mL` of `2.5xx10^(-5)M` of `Ba(OH_(2))`,To the resulting solution ,`99L` water is added ,calculate `pH` of final solution.Take `log0.303=-0.52`. |
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Answer» `[OH^(+)]_(f)=((500xx10^(-5))+(500xx2xx2.5xx10^(-5)))/(1000)=3xx10^(-5)M` `V_(1)=1L&V_(f)=100L` no of moles of `[OH^(+)]` in resulting solution =no. of moles of `[OH^(+)]` in final `3xx10^(-5)=[OH^(-)]_(f)xx100` `:.[OH^(-)]_(f)=3xx10^(-7)M( lt 10^(-6)M)` so `OH^(-)`ions coming form `H_(2)O` should also be considered. `H_(2)OhArrunderset(x) H^(+)+underset(x+3xx10^(-7))(OH^(-))` `K_(w)=x(x+3xx10^(-7))=10^(-14)` `,.x=((sqrt(13)-3)/(2))xx10^(-7)M=[H^(+)]` So `pH=7-log0.303=7.52` |
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| 430. |
`500mL` of `0.2M` aqueous solution of acetic acid is mixed with `500mL` of `0.2HCI` at `25^(@)C`. a. Calculate the degree of dissociation of acetic acid in the resulting solution and `pH` of the folution. b. If `6g` of `NaOH` is added to the above solution determine the final `pH.[K_(a)` of `CH_(3)COOH =2 xx 10^(-5)`. |
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Answer» (a) Meq. Of `CH_(3)COOH=500xx0.2=100` Meq. Of `HCI=500xx0.2=100` `:. [HCI]=(100)/(1000)=0.1, [CH_(3)COOH]=(100)/(1000)=0.1` For `CH_(3)COOH:` `{:(,CH_(3)COOHhArr,CH_(3)COO^(-)+,H^(+)),("Before dissociation",0.1,0,0.1("from HCI")),("After dissociation", (0.1x),x,(0.1+x)):}` Due to common ion effect, dissociation of `CH_(3)COOH` is very small in presence of HCI. Therefore `(0.1+x)=0.1 and (0.1-x)=0.1`. `:. K_(a)=(x xx0.1)/(0.1)` `:. x=K_(a)=1.75xx10^(-5)` Thus degree of dissociation `alpha= (x)/(0.1)=(1.757xx10^(-5))/(0.1)` `=1.75xx10^(-4)` `=0.000175=0.0175%` Also `[H^(+)]=0.1+x=0.1` , `(.: xltltalpha1)` `:. pH= -log[H^(+)]= -log[0.1]=1` (b) Meq. of NaOH mole of NaOH added `= (6)/(40)=0.15` Therefore new equilibrium will have `{:(CH_(3)COOH+,HCI+,NaOHrarr,CH_(3)COONa+,NaCI+,H_(2)O),(0.1,0.1,0.15,0,0,0),(0.5,0,0,0.05,0,0):}` Thus the solution will act as buffer having `[CH_(3)COOH]` `=(0.5)/(1000)` and `[CH_(3)COONa]=(0.5)/(1000)` Thus, `pH= -log K_(a)+log ((["Salt"])/(["Acid"]))` `= -log 1.75xx10^(-5)+log (([0.05//1000])/([0.05//1000]))` `pH=4.757` |
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| 431. |
`500mL` of `2 xx 10^(-3)M A1C1_(3)` and `500mol` of `4 xx 10^(-2)M` solution of `NaOH` are mixed and solution is diluted to `10^(-2)L` with water at room temperature wil a precipiate exist? Given: `K_(sp)`of `A1(OH)_(3) = 5 xx 10^(-33)`. |
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Answer» `[A1^(3+)]:` `{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before mixing"),,("After mixing"),,),(50xx2xx10^(-3),=,M_(2)xx1000,,):}` (Total volume `= 50 + 500 = 1000 mL)` `M_(2) = (500 xx 2xx 10^(-3))/(1000) = 10^(-3)M` It is further diluted to `10^(2)L`, so `[A1^(3+)] = (10^(-3))/(10^(2)L) = 10^(-5)M` Similarly `[overset(Theta)OH]: (4 xx 10^(-2))/(2xx10^(2)) = 2 xx 10^(-4)M` `{:(A1(OH)_(3)rarr,2A1^(3+)+,3overset(Theta)OH,,),(,10^(-5)+,(2xx10^(-4)),,):}` Hence, `Q_(sp) or I.P = [A1^(3+)] [overset(Theta)OH]^(3)` `= 10^(-5) xx (2xx 10^(-4))^(3) = 8 xx 10^(-17)`. `:. Q_(sp) gt K_(sp)` Hence, precipitation of `A1(OH)_(3)` occurs in solution. |
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| 432. |
The solubility product of a sparingly soluble salt `AX_(2)` is `3.2xx10^(-11)`. Its solubility (in `mo//L`) isA. `4 xx 10^(-4)`B. `5.6 xx 10^(-6)`C. `3.1 xx 10^(-4)`D. `2 xx 10^(-4)` |
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Answer» Correct Answer - D `AX_(3)hArr underset(s)(A^(2+))+underset(2s)(2X^(-))` `K_(sp)=[A^(2+)][X^(-)]^(2)` `3.2xx 10^(-11)=s(2s)^(2)` `4s^(3)=3.2 xx 10^(-11)` `s^(3)=8 xx 10^(-12)` `s=2 xx 10^(-4) mol L^(-1)` |
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| 433. |
Selective precipitation of ions in a mixture in the form of salts can be adding common ions gradually. Let us condsider selective precipitation of `CI^(-)` and `CrO_(4)^(2-)`ions in the form of AgCl and `Ag_(2)Cro_(4)` form a mixture having having 0.01M `CI^(-)` and 0.01M `CrO_(4)^(2-)`. For this, salt having `Ag^(+)` [like `AgNO_(3)(s)`] is added gradually. Given : `K_(sp) [AgCI]=10^(-10)` and `K_(sp)[Ag_(2)CrO_(4)]=4xx10^(-14)]` The percentage of one ion precipitated when another ion starts precipitation is : (Given: `(1)/(sqrt2)=0.7`)A. `98.7%`B. `99.7%`C. `97.3%`D. `92.7%` |
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Answer» Correct Answer - b |
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| 434. |
The rapid change of `pH` near the stoichiometric point of an acid-base titration is the basic of indicator detection. `pH` of the solution is related to the ratio of the concentration of conjugate acid `(Hin)` and base `(In^(-))` forms of the indicator by the expressionA. `log . ([ln^(-)])/([Hln])=pH-pK_(ln)`B. `log . ([ln^(-)])/([Hln])=pK_(ln)-pH`C. `log. ([Hln])/([ln^(-)])=pK_(ln)-pH`D. `log.([Hln])/([ln^(-)])=pH=pK_(ln)` |
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Answer» Correct Answer - A `Hln hArr H^(+) +ln^(-)` `K_(ln)=([H^(+)][ln^(-)])/([Hln])` `-logK_(ln)=-log[H^(+)]-log.([ln^(-)])/([Hln])` `pK_(ln)=pK_(a)-log.([ln^(-)])/([Hln])` `log.([ln])/([Hln])=pH-pK_(ln)` |
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| 435. |
The rapid change of `pH` near the stoichiometric point of an acid-base titration is the basic of indicator detection. `pH` of the solution is related to the ratio of the concentration of conjugate acid `(Hin)` and base `(In^(-))` forms of the indicator by the expressionA. `log.([In^(-)])/([HIn])=pK_(In)-pH`B. `log.([HIn])/([In^(-)])=pK_(In)-pH`C. `log.([HIn])/([In^(-)])=pH-pK_(In)`D. `log.([In^(-)])/([HIn])=pH-pK_(In)` |
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Answer» Correct Answer - D Acid indicatiors are generally weak acid. The dissociation of indicator Hin takes place as floows `HIn hArr H^(+)+In^(-)` `therefore K_(In)=([H^(+)][In^(-)])/([HIn])` or `[H^(+)]=K_(In).([HIn])/([In^(-)])` ….(i) `because pH=-log [H^(+)]` ...(ii) From eq. (i) and (ii) we get, `therefore pH=-log(K_(In).([HIn])/([In^(-)]))` `=-log K_(In)+log.([In^(-)])/([HIn])=pK_(In)+log.([In^(-)])/([HIn])` or `log.([In^(-)])/([HIn])=pH-pK_(In)` |
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| 436. |
The pH of a solution containing `NH_(4)OHandNH_(4)^(+)` ia 9. if `[NH_(4)^(+)]=0.1MandKa" of " NH_(4)^(+) " is "5xx10^(-10)` then what is `[NH_(4)OH]`?A. 0.05 MB. 20 MC. 0.5 MD. 2 M |
| Answer» Correct Answer - 1 | |
| 437. |
The solubility product of AgCl is `1.44 xx 10^(-4)` at `100^(@)C`. The solubility of silver chloride in boiling water may beA. `0.72 xx 10^(-4) M`B. `1.20 xx 10^(-2)M`C. `0.72 xx 10^(-2) M`D. `1.20 xx 10^(-4) M` |
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Answer» Correct Answer - B `AgCl hArr [Ag^(+)][Cl^(-)] , K_(sp) = S xx S, K_(sp) = S^(2)` `S = sqrt(K_(sp)) = sqrt(1.44 xx 10^(-4)) = 1.20 xx 10^(-2) M`. |
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| 438. |
0.3 g of `CH_(3) COOH` is dissolved in 100ml water to prepare a solution X. Now,0.2gm of NaOH is added in solution X to form solution Y. Finall, 0.245 g of `H_(2)SO_(4)` is added in Y to prepare solution Z,(Only Ist Hydrogen is ionizable). `CH_(3)COOH` behave like a weak acid with `K_(a)=10^(-5)`. [Moleucular weight :`CH_(3)COOH=60`,`NaOH=40,H_(2)SO_(4)=98]`[log7=0.85] What is the approximate pH of solution (Y) ?A. `5.15`B. `8.85`C. `8.15`D. `9.95` |
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Answer» Correct Answer - b |
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| 439. |
0.3 g of `CH_(3) COOH` is dissolved in 100ml water to prepare a solution X. Now,0.2gm of NaOH is added in solution X to form solution Y. Finall, 0.245 g of `H_(2)SO_(4)` is added in Y to prepare solution Z,(Only Ist Hydrogen is ionizable). `CH_(3)COOH` behave like a weak acid with `K_(a)=10^(-5)`. [Moleucular weight :`CH_(3)COOH=60`,`NaOH=40,H_(2)SO_(4)=98]`[log7=0.85] What is the pH of solution (Z) ?A. `5`B. `6`C. `8`D. `5` |
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Answer» Correct Answer - a |
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| 440. |
Which of the following aqueous solution will have a pH less than 7.0A. `KNO_(3)`B. NaOHC. `FeCl_(3)`D. NaCN |
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Answer» Correct Answer - C `Fe^(2+)` ions are hydrolysed to develop acidic nature. |
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| 441. |
An aquoeous solution twice alkaline as water, The pH of the solution is near toA. 6.7B. 6.4C. 7.3D. 7.6 |
| Answer» Correct Answer - 3 | |
| 442. |
Ostwald dilution law is applicable toA. Weak electrolytesB. Non-electrolytesC. Strong electrolytesD. All types of electrolytes |
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Answer» Correct Answer - A Ostwald dilution law is applicable to weak electrolytes |
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| 443. |
On the addition of a solution containing `CrO_(4)^(2-)` ions to the solution of `Ba^(2+), Sr^(2+)` and `Ca^(2+)` ions, the precipitate obtained first will be ofA. `CaCrO_(4)`B. `SrCrO_(4)`C. `BaCrO_(4)`D. Mixture of (a), (b), (c) |
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Answer» Correct Answer - C When we added barium ion in chromate ion solution we obtained yellow ppt of `BaCrO_(4)`. `BaCl_(2)+ K_(2)CrO_(4) rarr underset("Yellow ppt".)(BaCrO_(4)) darr + 2KCl` |
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| 444. |
Calculate `pH` of a solution whose `100ml` contains `0.2g NaOH` dissolved in it.A. 10.699B. 11.699C. 12.699D. 13.699 |
| Answer» Correct Answer - 3 | |
| 445. |
At `90^(@)` C, pH of an aq . Solution of a strong electrolyte is 7 . What is the nature of electrolyte ?A. neutralB. basicC. acidicD. none |
| Answer» Correct Answer - 2 | |
| 446. |
A solution cotaning `0.10 M` in `Ba(NO_(3))_(2)` and `0.10 M` in `Sr(NO_(3))_(2)`. If solid `Na_(2)CrO_(4)` is added to the solution, what is `[Ba^(2+)]`, when `SrCrO_(4)` beings to precipitate? `[K_(sp)(BaCrO_(4)) = 1.2 xx 10^(-10), K_(sp) (SrCrO_(4)) = 3.5 xx 10^(-5)]`A. `7.4 xx 10^(-7)`B. `2.0 xx 10^(-7)`C. `6.1 xx 10^(-7)`D. `3.4 xx 10^(-7)` |
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Answer» Correct Answer - D `K_(sp)(SrCrO_(4)) = [Sr^(2+)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)] = (3.5 xx 10^(-5))/(0.1) = 3.5 xx 10^(-4)` `K_(sp)(BaCrO_(4)) = [Ba^(2)][CrO_(4)^(2-)]` `[CrO_(4)^(2-)]_("total") = [CrO_(4)^(2-)]` from `SrCrO_(4)` `[Ba^(2+)] = (1.2 xx 10^(-10))/(3.5 xx 10^(-4)) = 3.4 xx 10^(-7)` |
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| 447. |
For a solution of weak triprotic acid `H_(3)A(K_(a_(1))gtgtK_(a_(2)),K_(a_(3)):K_(a_(2))=10^(-8),K_(a_(3))=10^(-13),[A^(3-)]=10^(-17)M`.Determine `pH` of solutions. Report your answer as `0` if you find data insufficient. |
| Answer» `{:(,HA^(2-),hArr,H^(+),+,A^(3-),,,K_(a_(3))=10^(-13)),(t=eq,,10^(-8),,,,10^(-17),):}` | |
| 448. |
Which of the following is a strong electrolyte?A. `NH_(3)`B. `Ca(OH)_(2)`C. `BaCl_(2)`D. `H_(3)PO_(4)` |
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Answer» Correct Answer - C `BaCl_(2)` is strong electrolyte |
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| 449. |
What is the `pH` of `7.0 xx 10^(-8)M` acetic acid. What is the concentration of un-ionsed acetic acid. `K_(a)` of `CH_(3)COOH = 1.8 xx 10^(-5)`. |
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Answer» a. In such dilute solutions, complete ionsiation of weak acid occurs, but ionisation of water must also be considered. Let `x [oversetΘ)OH] then [H_(3)O^(o+)] = x +(7.0 xx 10^(-8))` (From `H_(2)O` and acetic acid) `:.K_(w) - [H_(3)O^(o+)] [overset(Θ)OH] rArr (x +7.0 xx 10^(-8)) (x) = 10^(-14)`.or `rArr x^(2) +(7.0 xx 10^(-8))x - 10^(-14) = 0` `rArr x = 7.1 xx 10^(-8)M = [overset(Θ)OH]` Hence, `[H_(3)O^(o+)] (7.1 xx 10^(-8) +7.0 xx10^(-8))` `= 1.4 xx 10^(-7)M` `pH = - log (1.4 xx 10^(-7)) = 6.85` b. From charge balance (electroneutrality) Total negative charge = Total positive charge `[CH_(3)COO^(Θ)] +[overset(Θ)H] = [H_(3)O^(o+)]` `:. [CH_(3)COO^(Θ)] = [H_(3)O^(o+)] - [overset(Θ)OH]` `= (1.4 xx 10^(-7) - (7.1 xx 10^(-8))` c. `CH_(3)COOH +H_(2)O hArr CH_(3)COO^(Theta) +H_(3)O^(oplus)` `K_(a) = =([CH_(3)COO^(Θ)][H_(3)O^(o+)])/([CH_(3)COOH]) = 1.8 xx 10^(-15)` `[CH_(3)COOH] = ([CH_(3)COO^(Θ)][H_(3)O^(o+)])/(K_(a))` `= ((7 xx 10^(-8))(1.4 xx 10^(-7)))/(1.8xx10^(-5))` `= 5 xx 10^(-10)M` |
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| 450. |
`H_(3)A` is a weak triprotic acid `(K_(a1)=10^(-5),K_(a2)=10^(-9),K_(a3)=10^(-13)` What is the value of pX of 0.1 M `H_(3)A` (aq.) solution ? Where pX=-log X and X=`[[A^(3-)]]/[[HA^(2-)]]`A. `7`B. `8`C. `9`D. `10` |
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Answer» Correct Answer - D `alpha` is negligible w.r.t `1` `[H^(+)]` mainly from first step `[H^(+)] = sqrt(K_(a) xx C) = sqrt(10^(-5) xx 0.1) = 10^(-3)` `:. K_(a_(3)) = ([H^(+)][A^(3-)])/([HA^(2-)]) = 10^(-13) = (10^(-3)[A^(3-)])/([HA^(2-)])` `X = ([A^(3-)])/([HA^(2-)]) = 10^(-10)` , `:. pX = 10` |
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