Question

Calculate the pH of a buffer solution made from equal volumes of 0.3M HC2H3O2 and 0.6M NaC2H3O2. (Ka = 1.8 x 10-5)

Answer 1

Answer:

pH of buffer is 5.05

Explanation:

Let's assume that V mL of both solutions were added.

Number of moles of $HC_{2}H_{3}O_{2}$ added = $\frac{0.3\times V}{1000}moles$

Number of moles of $C_{2}H_{3}O_{2}^{-}$ added = $\frac{0.6\times V}{1000}moles$

According to Henderson-Hasselbalch equation for a buffer consisting of an weak acid ($HC_{2}H_{3}O_{2}$) and it's conjugate base ($C_{2}H_{3}O_{2}^{-}$)-

$pH=pK_{a}(HC_{2}H_{3}O_{2})+log[\frac{moles of C_{2}H_{3}O_{2}^{-}}{moles of HC_{2}H_{3}O_{2}}]$

$pK_{a}=-log(K_{a})$

So, $pH=-log(1.8\times 10^{-5})+log[\frac{\frac{0.6\times V}{1000}}{\frac{0.3\times V}{1000}}]$

So, $pH=-log(1.8\times 10^{-5})+log[\frac{0.6}{0.3}]$

So, pH = 5.05