Question

Acetyl salicylic acid (aspirin) ionises in water as: $H{C}_9{H}_7{O}_4+H_{2}O\to H_3{O}^{+}+C_9{H}_7{O}_4^{-};$ $(K_a=2.75\times {10}^{-9})$. If two tablets of aspirin each of 0.32 g is dissolved in water to produce 250 mL solution, calculate $\left[\stackrel{⊝}{O}H\right].$

• A. $1.61\times {10}^{-9}M$
• B. $1.61\times {10}^{-7}M$
• C. $1.61\times {10}^{-3}M$
• D. $1.61\times {10}^{-5}M$

Answer 1

The correct option is A $1.61\times {10}^{-9}M$

$\left[H{C}_9{H}_7{O}_4\right]=\frac{0.32\times 2\times 1000}{180\times 250}=0.014M$
$\because \alpha =\sqrt{\frac{K_a}{C}}=\sqrt{\frac{2.75\times {10}^{-9}}{0.014}}=4.43\times {10}^{-4}$

$\therefore \left[H^{\oplus }\right]=C.\alpha =0.014\times 4.43\times {10}^{-4}=6.21\times {10}^{-6}M$

$\left[C_9{H}_7{O}_4^{\ominus }\right]=C.\alpha =6.21\times {10}^{-6}M$

$\left[O{H}^{⊝}\right]=\frac{{10}^{-14}}{\left[H^{\oplus }\right]}=\frac{{10}^{-14}}{6.21\times {10}^{-6}}=1.61\times {10}^{-9}M$