Question

Calculate the ratio of $C{H}_{3}N{H}_2$ to $C{H}_{3}N{H}_3^{+}C{l}^{-}$ required to create a buffer with pH = 10.14
$K_{b}ofC{H}_{3}N{H}_2=4.4\times {10}^{-4}$
take $log2.27=0.356,log0.313=-0.504$

• A. 0.5
• B. 1.05
• C. 0.31
• D. None of the above

Answer 1

The correct option is C 0.31

Given, $K_{b}ofC{H}_{3}N{H}_2=4.4\times {10}^{-4},pH=10.14$
We need the $K_a$ of the methylammonium ion:
We know at ${25}^{o}C{K}_a\times K_b={10}^{-14}$
$K_a\text{for}C{H}_{3}N{H}_3^{+}C{l}^{-}={10}^{-14}/4.4\times {10}^{-4}=2.27\times {10}^{-11}$
Using the Henderson-Hasselbalch equation:
$pH=p{K}_a+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
Substitute and solve for the base/acid ratio:
$10.14=-log2.27\times {10}^{-11}+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
$10.14=11-0.356+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
$10.14=10.644+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
$log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]=-0.504$
$\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}=antilog(-0.504)$
$\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}=0.313$