Question

There is a weak base 'B' having $\text{pK}_b = 5.691$ of molarity 0.02M. When 0.02M HCl solution has been added, then pH of resultant buffer solution has been found to be 9. Take total volume of resultant buffer solution to be 100 ml. Find the value of 'x' & 'y', where 'x' is volume of HCl solution in ml & 'y' is volume of 'B' solution in ml. Given $\log(5) = 0.691$

• A. x=14.29, y=85.71
• B. x=15, y=85
• C. x=20, y=80
• D. x=40, y=60

Hint:

For buffer problems, determining the ratio of conjugate pair from the pH equation allows you to relate the reactant volumes directly.

Answer 1

The Correct Option is A

Total volume $x + y = 100$.
Moles of Base $B = 0.02y$ millimoles.
Moles of Acid $HCl = 0.02x$ millimoles.
Reaction: $B + H^+ \to BH^+$.
For a buffer (pH 9, basic), B must be in excess.
Salt formed ($BH^+$) $= 0.02x$.
Base remaining $= 0.02y - 0.02x = 0.02(y-x)$.
Buffer Formula: $\text{pOH} = \text{pK}_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$.
$\text{pH} = 9 \implies \text{pOH} = 14 - 9 = 5$.
$5 = 5.691 + \log \frac{0.02x}{0.02(y-x)}$.
$-0.691 = \log \frac{x}{y-x}$.
Given $\log 5 = 0.691$, so $-0.691 = \log (1/5)$.
$\frac{x}{y-x} = \frac{1}{5}$.
$5x = y - x \implies y = 6x$.
Substitute into volume equation: $x + 6x = 100 \implies 7x = 100 \implies x = 14.285$.
$y = 100 - 14.285 = 85.715$.