Question

When $0.1$ mol of $C{H}_{3}N{H}_2(K_b=5\times {10}^{-4})$ is mixed with $0.08$ mol of $HCl$ and diluted to $1L$, what will be the $H^{\oplus }$ concentration in the solution?

• A. $8\times {10}^{-2}M$
• B. $8\times {10}^{-11}M$
• C. $1.6\times {10}^{11}M$
• D. $8\times {10}^{-5}M$

Answer 1

Answer: (B)

$8\times {10}^{-11}M$

As we know,

$C{H}_{3}N{H}_2+HCl\to C{H}_{3}N{H}_3^{+}+C{l}^{-}$

Initial $0.1$ $0.08$ $-$

Final $0.02$ $-$ $0.08$

$pOH=p{K}_b+log\frac{\left[salt\right]}{\left[base\right]}=4-log5+log\frac{0.8}{0.2}=4-log5+log4=3.903$

So, $pH=14-pOH=10.09$ and $\left[H^{+}\right]={10}^{-pH}=8\times {10}^{-11}M$