Question

The base imidazole has a $K_b$ of $9.8\times {10}^{-8}$ at ${25}^{\circ }$ C.

In what ratio of volumes should 0.02 M HCl and 0.02 M imidazole be mixed to make 100 ml of a buffer at pH 7 ?

• A. $\frac{3}{5}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{1}$

Answer 1

The correct option is C

$\frac{1}{2}$

As pH = 7, pOH = 14 – 7 = 7 (at ${25}^{\circ }$ C).

$p{K}_b=-log{K}_b=-log(9.8\times {10}^{-8})$ = 7.0088

Applying

pOH = $p{K}_b$ + log $\frac{\left[Salt\right]}{\left[Base\right]}$

7 = 7.0088 + log $\frac{\left[Salt\right]}{\left[Base\right]}$

log $\frac{\left[Salt\right]}{\left[Base\right]}$ = - 0.0088

Taking antilog, $\frac{\left[Base\right]}{\left[Salt\right]}$ = 1.019

Or $\frac{millimoleofbase}{millimoleofsalt}$ = 1.019 ... (1)

Suppose $V_1$ mole of HCI is mixed with $V_2$ ml of imidazole (base) to make the buffer. millimole of HCl = 0.02$V_1$ or millimole of imidazole = 0.02$V_2$

As the buffer is of the base and its salt, 0.02 millimole of HCl will combine with 0.02 millimole of base to give 0.02 millimole of salt.
millimole of salt = millimole of HCl = 0.02 $V_1$
and millimole of base left = 0.02$V_2$ – 0.02$V_1$

From (1), we get, $\frac{0.02(V_2–V_1)}{0.02V_1}$ = 1.019

Or $\frac{(V_2–V_1)}{V_1}$ = 1.019 ... (2)
Given that $V_1$ + $V_2$ = 100 ... (3)
From (2) and (3) we get, V₁ = 33 mL
and $V_2$ = 67 mL
$\frac{V_1}{V_2}=\frac{33}{67}=\frac{1}{2}$