Questions on Chemical Kinetics – subjective & objective
Chemical Kinetics (Subjective Unsolved)
- 25 mL of a weak base BOH was titrated with 0.5 M HCl. The pH of solution upon addition of 10 mL acid was 8.6 and that upon addition of 25 mL acid was 8. Calculate pH solution when (i) o mL (ii) 20 mL and (iii) 50 mL of acid has been added.
- 12.5 mL solution of 0.5 M H3XO4is titrated with 0.5M NaOH. Calculate pH at the following stages.
- a) When 1st step neutralisation is complete
- b) When 2nd step neutralisation is complete.
- c) When 3rd step neutralisation is complete.
K1, K2 and K3 for H3XO4 are 1.0 ´ 10–4, 2.0 ´ 10–7 and 4.0 ´ 10–11 respectively.
- An acid-base indicator which is actually a weak monobasic acid exists 50% dissociated in an acid solution of pH = 4.60. The acid colour and basic colour of the indicator are quite distinct and say they are colour A and colour B respectively. Colour A predominates over colour B when concentration of acidic form of the indicator is at least 10 times that of the conjugate base of the indicator. However, colour B predominates over colour A only when concentration of basic form is about 20 times be a suitable one for indicating the end point of a titration carried out with a mixture of two weak acids (20 mL 0.1 M HA + 30 mL 1.5 M HB) vs. 0.5 M NaOH. Ka of HA and that of HB are 1 ´ 10–5 and 2 ´ 10–10 respectively.
- When 20 mL solution f 4 M NaHCO3 solution is mixed with 100 mL sample of blood containing 12.4 g of H2CO3 the pH of the mixture was found to be 6.4 What volume of 5 M NaHCO3 solution should be mixed with a 10 mL of the same blood sample which is 1.0 M in H2CO3 in order to maintain a pH of 8.0.
- 50 mL of a weak monobasic acid was titrated with 0.2M NaOH and volume of alkali needed for the end point was 20 mL. The pH of solution upon addition of 12 mL alkali was 5.16. 100 mL solution of this acid at concentration of 0.2 M and also containing M3+ ion at a given of 0.1 M should be mixed with what volume of 0.5 M NaOH so as to produce a buffer solution in which metal hydroxide may begin to precipitate?
- In pure formic acid following auto ionisation equilibrium exists
HCOOH(l) + HCOOH(l) HCOOH2+ + HCOO–
If the formic acid at 27°C is 0.004$ auto ionised. Calculate rate constant of the nitro molecular proton transfer of formic acid, when the rate constant of proton transfer between conjugate acid of formic acid and its conjugate base is 2 ´ 102 M2. (Density of formic acid = 1.22 g/mL).
- a = = 4 ´ 10–5
[HCOOH2+] = [HCOO–] = Ca = 26.5 ´ 4 ´ 105 = 1.06 ´ 10–3
K = 1.06 ´ 10–3 ´ 1.06 ´ 1.06 ´ 10–3
K = \ K1 = K ´ K2 = 1.12 ´ 10–6 ´ 2 ´ 102 = 2.24 ´ 10–4
- The average concentration of SO2 in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298K. Given that solubility of SO2 in water at 298K is 1.3653 mol L–1 and the pKa of H2SO3 is 1.92, estimate the pH of rain on that day.
- A strong electrolyte in solution remains
(A) Completely dissociated in solution at all dilute
(B) Completely ionised but no completely dissociated
(C) Partially ionised and partially dissociated
(D) Completely ionised and completely dissociated at all concentrations
- A pure monprotic polar solvent (MW = 50) at 25°C exists 2 ´ 10–6% dissociated. The density of the pure solvent is 1.28/mL. The Ka of the solvent is
(A) 9.6 ´ 10–15 (B) 2.0 ´ 10–15
(C) 4 ´ 10–16 (D) None of these
- The species that is not present in the aqueous solution of HCl is
(A) Cl– (B) H+
(C) OH– (D) H2O
- An indicator exists 80% ionised in a solution of pH = 6.6. The acid colour of the indicator predominates over the basic colour when concentration of acid is 8 times that of its conjugate base. The basic colour of the indicator prevails over its acidic colour when the concentration of basic form is 16 times that of the acid. The colour of acid is A, that of its conjugate base is B and the intermediate colour of A and B is C. Thus, in pure water at 25°C, the indicator will display.
(A) Colour A (B) Colour B
(C) Colour C (D) None of these
- The concentration of M2+ ion (atomic mass = 25) in a saturated aqueous solution of M(OH)2 is 100 ppm. The Kspof M(OH)2 is
(A) 2.85 ´ 10–10 (B) 2.38 ´ 10–9
(C) 2.56 ´ 10–7 (D) None of these
- The pH of a solution of 0.5M NaHCO3 (K1 and K2 of H2CO3 being 2 ´ 10–7 and 4 ´ 10–11 respectively) is
(A) 9.85 (B) 7.9
(C) 10.02 (D) 18.55
- A certain weak acid has a dissociation constant 1 ´ 10–5. The equilibrium constant for its reaction with a strong base is:
(A) 1.0 ´ 10–5 (B) 1.0 ´ 10–9
(C) 1.0 ´ 1010 (D) 1.0 ´ 109
- The solubility of I2 in water may be increased by adding
(A) KI (B) NaCl
(C) CHI3 (D) K4Fe(CN)6
- The pH of 0.1M solution of the following salts/acids increase in the order:
(A) NaCl < NH4Cl < NaCN < HCl (B) HCl < NH4Cl < NaCl < NaCN
(C) NaCN < NH4Cl < NaCl < HCl (D) HCl < NaCl < NaCN < NH4Cl
- Which of the following statements are correct:
(A) The pH of 10–9M HCl is less than 7 at 25°C
(B) The conjugate base of H2S is S2–
(C) Autoprotolysis constant of water increases with temperature
(D) When a solution of weak monoprotonic acid is titrated against a strong base, at half neutralisation.
Problem 1 The pH blood is maintained by proper balance of H2CO3 and NaHCO3 concentrations. What volume of 5 M NaHcO3 solution should be mixed with a 10 mL sample of blood which 2 M in H2CO3 in order to maintain a pH of 7.4?
Ka(H2CO3) = 7.8 ´ 10–7.
Solution: Let x ml 5 M NaMCO3 be added
\ No. of millimole of NaHCO3 added = 5x
No. of millimole of H2CO3 taken = 10 ´ 2 = 20
Ka = 7.8 ´ 10–7
\ pKa = 6.01
For acidic buffer mixture
pH = pKa +
7.4 = 6.10 +
\ x = 80
So volume of 5 M NaHCO3 solution to be added is 80 mL
Problem 2: A certain acid base indicator is red in acid solution and blue in basic solution. At pH = 5, 75% of the indicator is present in the solution in its blue form. Calculate Ka for the indicator and pH range over which the indicator changes from 90% red 10% blue to 90% blue – 10% red.
Solution: HIn In– + H+
Ka = 10–5 ´ = 3 ´ 10–5
The above equation, after taking negative log of both sides may be put as
pH = pKa + log
pH for 90% red – 10% blue is as calculated below
pH = – log (3 ´ 10–5) + log
= 4.52 – 0.95 = 3.47
pH for 10% red – 90% blue is as calculated below
pH = 4.52 +
= 4.52 + 0.95
Thus the required pH range is 3.47 – 5.47
Problem 3: Let the solubilities of BaSO4 and BaCrO4 be x and y mol L–1 negatively
BaSO4(s) Ba2+ + SO42–
x + y x
BaCrO4(s) Ba2+ + CrO42–
x + y y
\ y = 2.4x
Putting the value of y
1 µ 10–10 = (x + 2.4x)x = 3.4x2
x = = 542 ´ 10–6 mol L–1
Putting the value of x
y = 2.4 ´ 5.42 ´ 10–6 = 1.3 ´ 10–5 mol L–1
Problem 4: 20 mL of 0.2 M NaOH is added to 50 mL of 0.2 M acetic acid. What is the pH of the solution? Calculate the additional volume of 0.2 M naOH required to make pH of the solution 4.74. Ka of CH3COOH = 1.8 ´ 10–5
Solution: Acetic acid taken = 5.0 ´ 0.2 = 10 millimole
NaOH added = 20 ´ 0.2 = 4 millimole
Salt formed = 4 millimole
Acid remaining free = 10 4 = 6 millimole
Total volume of solution = 20 + 50 = 70 mL
pKa= –log(1.8 ´ 10–5) = 4.74
Resulting mixture is an acid buffer mixture for which
pH = pKa +
= 4.74 +
= 4.74 + = 4.57
To make the pH of solution equal to 4.74, we have to make [salt] = [acid]. To do so 5 mL 0.2 M NaOH 1 millimole will have so be added further.
Problem 5: If at 25°C, the equivalence point of a titration of 50 mL of a solution of a weak monobasic acid occurs when 30 mL of 0.1 M NaOH has been added and the pH of solution is 6.0 after the addition of 18 mL of NaOH solution. Calculate Ka for the acid.
Solution: Since weak acid is monobasic and NaOH is also monoacidic so they must react in 1:1 molar proportion.
Amount of NaOH added for the end point = 30 ´ 0.1 = 3 millimole. So, amount of acid present in 50 mL solution = 3 millimole. During titration when 18 mL 0.1 i.e. 1.8 millimole NaOH is added, the salt formed will be 1.8 millimole and acid remaining free will be 3 – 1.8 i.e. 1.2 millimole. The mixture will be an acid buffer mixture for which
pH = pKa +
6.0 = pKa +
\ pKa =
Problem 6: Calculate the solubility of AgCN in a buffer solution of pH = 3.0. Ksp (AgCN) = 1.2 ´ 10–16‑, Ka (HcN) = 4.8 ´ 10–10. There is no CN– or Ag+ in the buffer previously.
Solution: AgCN(s) Ag++ CN– Ksp= 1.2 ´ 10–16
CN–+ H+ HCN
AgCN(s) + H+ Ag+ + HCN K = = 2.5 ´ 10–7
10–3 x x
\ = 2.5 ´ 10–7
\ x = = 1.58 ´ 10–5M
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