Question

What is $\left[B^{-}\right]$ in a solution that has 0.03 M HA and 0.1 M HB?

$K_a$ for HA and HB are $1.38\times {10}^{-4}$ and $1.05\times {10}^{-10}$, respectively.

• A. $5.15\times {10}^{-9}M$
• B. $5.15\times {10}^{-3}M$
• C. $5.15\times {10}^{-5}M$
• D. $5.15\times {10}^{-7}M$

Answer 1

Answer: (C)

$5.15\times {10}^{-5}M$

$HA\rightleftharpoons H^{\oplus }+A^{⊝};K_a=1.38\times {10}^{-4}$
$HB\rightleftharpoons H^{\oplus }+B^{⊝};K_a=1.05\times {10}^{-10}$
It is thus evident on account of low $K_a$ values for HB, $\left[H^{\oplus }\right]$ from HB is appreciably small in comparison to $\left[H^{\oplus }\right]$ from HA and thus may be neglected.
For $HA:K_a=\frac{\left[H^{\oplus }\right]\left[A^{⊝}\right]}{\left[HA\right]}$
$1.38\times {10}^{-4}=\frac{[H^{\oplus }{]}^2}{\left[HA\right]}$
$[H^{\oplus }{]}^2=1.38\times {10}^{-4}\times [HA]$
$=1.38\times {10}^{-4}\times 0.03$
Therefore, $\left[H^{\oplus }\right]=\left[A^{⊝}\right]=2.04\times {10}^{-3}M$
Now for $HB:K_a=\frac{\left[H^{\oplus }\right]\left[B^{⊝}\right]}{\left[HB\right]}$
$1.05\times {10}^{-6}=\frac{2.04\times {10}^{-3}\times \left[B^{⊝}\right]}{0.1}$
$\left[B^{⊝}\right]=5.15\times {10}^{-5}M$