# What molar ratio of sodium acetate to acetic acid should be used to prepare a buffer with pH = 4.5? K_a acetic acid = 1.8 x 10^-5?

Jan 10, 2017

$\left(\left[\text{AcO"^-])/(["HOAc}\right]\right) = 1.76$

#### Explanation:

$p H = p {K}_{a} - {\log}_{10} \left(\left(\left[\text{AcO"^-])/(["HOAc}\right]\right)\right)$

And thus,

${\log}_{10} \left(\left(\left[\text{AcO"^-])/(["HOAc}\right]\right)\right) = p {K}_{a} - p H = - {\log}_{10} \left(1.8 \times {10}^{-} 5\right) - 4.5 = 4.75 - 4.5 = 0.245$

And thus ${10}^{0.245} = \frac{\left[A c {O}^{-}\right]}{\left[H O A c\right]} = 1.76$