Question

$20$ mL of $0.1M$ weak acid $HA(K_a={10}^{-5})$ is mixed with the solution of $10$ mL of $0.3MHCl$ and $10$ mL of $0.1MNaOH$. Find the value of $\frac{\left[A^{-}\right]}{\left(\right[HA]+[A^{-}\left]\right)}$ in the resulting solution?

• A. $2\times {10}^{-4}$
• B. $2\times {10}^{-5}$
• C. $2\times {10}^{-3}$
• D. $0.05$

Answer 1

The correct option is A $2\times {10}^{-4}$

As we know,
$C\alpha =\left[A^{-}\right]=\left[H^{+}\right]=\sqrt{K_a\times C}={10}^{-3}$ M.
So, $20$ mL of solution will have $\left[A^{-}\right]=20\times {10}^{-3}\times {10}^{-3}$.
Since $\left[HA\right]>>>\left[A^{-}\right]$, $\frac{\left[A^{-}\right]}{\left[HA\right]+\left[A^{-}\right]}=\frac{\left[A^{-}\right]}{\left[HA\right]}=\frac{2\times {10}^{-5}}{0.1}=2\times {10}^{-4}.$