NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Ionic Equilibrium
NTA Abhyas 2020 · Chemistry · Ionic Equilibrium
K_ a for HCN is 5 x 10^ - 10 at 25^ o C. For maintaining a constant...
K_{a} for HCN is 5 x 10^{- 10} at 25^{o} C. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 mL of 2 M HCN solution is:
Options
- A4 mL
- B7.95 mL
- C9.3 mL
- D2 mL
Correct answer
D. 2 mL
Step-by-step solution
On mixing the given solutions, acidic buffer will be formed. Thus applying Henderson equation to find the pH-
\text{pH} = - \text\log 5 \times 1 {0}^{- 1 0} + \text\log \dfrac{5 \times \text{V} / \text{V} + 1 0}{1 0 \times 2 / \text{V} + 1 0}
9 = - \text\log 5 \times 1 {0}^{- 1 0} + \text\log \dfrac{\text{V}}{4}
9 = 1 0 - 0 \text{.} 7 + \text\log \dfrac{\text{V}}{4}
- 0 \text{.} 3 = \text\log \dfrac{\text{V}}{4}
∴\text\log \dfrac{\text{V}}{4} {=}^{-} 0 \text{.} 3 = \text\log \dfrac{1}{2}
&
V = 2 mL.
\text{pH} = - \text\log 5 \times 1 {0}^{- 1 0} + \text\log \dfrac{5 \times \text{V} / \text{V} + 1 0}{1 0 \times 2 / \text{V} + 1 0}
9 = - \text\log 5 \times 1 {0}^{- 1 0} + \text\log \dfrac{\text{V}}{4}
9 = 1 0 - 0 \text{.} 7 + \text\log \dfrac{\text{V}}{4}
- 0 \text{.} 3 = \text\log \dfrac{\text{V}}{4}
∴\text\log \dfrac{\text{V}}{4} {=}^{-} 0 \text{.} 3 = \text\log \dfrac{1}{2}
&
V = 2 mL.