# What is the pH of a buffer that was prepared by adding 3.96 g of sodium benzoate, NaC_7H_5O_2, to 1.00 L of 0.0100 M benzoic acid, HC_7H_5O_2?

## Assume that there is no change in volume. The ${K}_{a}$ for benzoic acid is $6.3 \cdot {10}^{-} 5$.

Mar 14, 2016

The $\text{pH}$ of the buffer is 4.64.

#### Explanation:

Let's represent the benzoate ion, $\text{C"_7"H"_5"O"_2^"-}$, as $\text{A"^"-}$.

Then the equation for the dissociation of benzoic acid becomes

$\text{HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-}$; ${K}_{\text{a" = 6.3 ×10^"-5}}$

The typical formula for calculating the $\text{pH}$ of a buffer is the Henderson-Hasselbalch equation:

$\textcolor{b l u e}{| \overline{\underline{\text{pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]}}} |}$

Our job is to calculate the numbers to insert into the formula.

(a) "p"K_"a" = "-"logK_"a" = "-"log(6.3 × 10^"-5") = 4.20

(b) $\left[{A}^{\text{-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O}} _ 2\right]$

3.96 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2))) × ("1 mol NaC"_7"H"_5"O"_2)/( 144.10 color(red)(cancel(color(black)("g NaC"_7"H"_5"O"_2)))) = "0.027 48 mol NaC"_7"H"_5"O"_2

[A^"-"] = "0.027 48 mol"/"1 L" = "0.027 48 mol/L"

(c) $\text{[HA] = 0.0100 mol/L}$

Now, we insert these values into the Henderson-Hasselbalch equation.

$\text{pH"= "p"K_"a" + log (("[A"^"-""]")/"[HA]") = 4.20 + log(("0.027 48" color(red)(cancel(color(black)("mol/L"))))/ (0.0100 color(red)(cancel(color(black)("mol/L"))))) = 4.20 + log(2.748) = 4.20 + 0.439 = 4.64}$

The $\text{pH}$ of the buffer is 4.64.