# Question #2c8ae

Dec 10, 2015

$\left[{H}^{+}\right] = 7.58 \times {10}^{- 6} M$

#### Explanation:

We console this question using Henderson-Hasselbalch equation:

$p H = p {K}_{a} + \log \left(\frac{\left[{C}_{2} {H}_{3} {O}_{2}^{-}\right]}{\left[H {C}_{2} {H}_{3} {O}_{2}\right]}\right)$

$p {K}_{a} = - \log {K}_{a} = - \log \left(1.8 \times {10}^{- 5}\right) = 4.74$

$\implies p H = 4.74 + \log \left(\frac{0.48}{0.2}\right) = 5.12$

Now we can calculate the concentration of ${H}^{+}$ by:

$\left[{H}^{+}\right] = {10}^{- p H} = {10}^{- 5.12} = 7.58 \times {10}^{- 6} M$

Here is a video that explains Henderson-Hasselbalch equation:
Acid - Base Equilibria | Buffer Solution.