Question

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 * 10^-5`.

Answer 1

The `"pH"` of the buffer is 4.64.

Explanation:

Let's represent the benzoate ion, `"C"_7"H"_5"O"_2^"-"`, as `"A"^"-"`.

Then the equation for the dissociation of benzoic acid becomes

`"HA" + "H"_2"O" ⇌ H_3"O"^+ + "A"^"-"`; `K_"a" = 6.3 ×10^"-5"`

The typical formula for calculating the `"pH"` of a buffer is the Henderson-Hasselbalch equation:

`color(blue)(|bar(ul("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) `[A^"-"] = ["C"_7"H"_5"O"_2^"-"] = ["NaC"_7"H"_5"O"_2]`

`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) `"[HA] = 0.0100 mol/L"`

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

`"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 `"pH"` of the buffer is 4.64.