# Calculate pH of buffer solution with 0.1 M each of CH_3COOH & CH_3COONa. What will be the change in pH on addition of the following: ? (pK_a = 1.8 xx 10^-5). The total volume of solution is 1 L.

## (i) $0.01$ $M$ $H C l$ (ii) $0.01$ $M$ $N a O H$

Jun 22, 2017

The buffer equation holds that.........

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

#### Explanation:

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

Where $H A$ is the parent acid; and the $\left[{A}^{-}\right]$ is its conjugate. For the derivation of this formula, which an undergraduate should be able to perform, see here.

And thus for this problem..........

$p H = 4.76 + {\log}_{10} \left\{\frac{\left[A c {O}^{-}\right]}{\left[H O A c\right]}\right\}$

And note that $4.76 = - {\log}_{10} \left(1.8 \times {10}^{-} 5\right)$

Now the starting conditions specify that there are equal concentrations of acetate and free acid, and thus $p H = p {K}_{a} = 4.76$. Why? Because ${\log}_{10} 1 = 0$, so given the equation, $\left[A c {O}^{-}\right] = \left[H O A c\right]$, and $\frac{\left[A c {O}^{-}\right]}{\left[H O A c\right]} = 1$.

Anyway, work thru the problems, and post them back in this thread. The $p H$ will be round about $4.76$, i.e. the $p {K}_{a}$ of the parent acid.