Question

Equilibrium constant for the following reaction is $1\times {10}^{-9}$:

$C_5{H}_{5}N(aq.)+H_{2}O\left(l\right)\rightleftharpoons C_5{H}_5{NH}^{+}(aq.)+{OH}^{-}(aq.)$

Determine the mole of pyridinium chloride $(C_5{H}_{5}N.HCI)$ that should be added to 500 mL solution of 0.4 M pyridine $\left(C_5{H}_{5}N\right)$ to obtain a buffer solution of pH = 5:

• A. 0.1 mole
• B. 0.2 mole
• C. 0.3 mole
• D. 0.4 mole

Answer 1

The correct option is B 0.2 mole

$C_6{H}_{5}N\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C_6{H}_{5}NH\pm \left(aq\right)+O{H}^{-}\left(aq\right)k={10}^{-9}$

$pH=5$

we know for buffer

$pH=p{K}_a-\log \frac{\left[base\right]}{\left[acid\right]}$

$\Rightarrow 5=14-p{K}_b+\log \frac{\left[pyride\right]}{\left[C_6{H}_{5}NHCl\right]}$

$\Rightarrow 5=10-9+\log \frac{0.4}{x}\times 0.5$

$\Rightarrow \log \frac{0.2}{x}=0$

$\Rightarrow \frac{0.2}{x}=1\Rightarrow x=0.2$

$\therefore 0.2mol$