# What is the pH when 0.600 moles of potassium benzoate (KC_7H_5O_2) has been added to 1.50 of water?

Sep 9, 2016

The pH is 8.90.

#### Explanation:

Our first task is to calculate the concentration of the potassium benzoate,

$\text{Molarity" = "moles"/"litres" = "0.600 mol"/"1.50 L" = "0.400 mol/L}$

The reaction is

$\text{C"_7"H"_5"O"_2^"-" + "H"_2"O" ⇌ "HC"_7"H"_5"O"_2 + "OH"^"-}$

For benzoic acid, K_a = 6.25 × 10^"-5".

For the benzoate ion, as above, ${K}_{\text{b" = K_"w"/K_"a" = (1.00 × 10^"-14")/(6.25 × 10^"-5") = 1.60 × 10^"-10}}$

Now, we can set up an ICE table to calculate the concentrations.

$\textcolor{w h i t e}{m m m m m m} \text{C"_7"H"_5"O"_2^"-" + "H"_2"O" ⇌ "HC"_7"H"_5"O"_2 + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.400 \textcolor{w h i t e}{m m m m m m m m m} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1": color(white)(mm)"-} x \textcolor{w h i t e}{m m m m m m m m l l} + x \textcolor{w h i t e}{m m} + x$
$\text{E/mol·L"^"-1": color(white)(ll)"0.400 -} x \textcolor{w h i t e}{m m m m m m m m l} x \textcolor{w h i t e}{m m m l} x$

${K}_{\text{b" = (["HC"_7"H"_5"O"_2]["OH"^"-"])/(["C"_7"H"_5"O"_2^"-"]) = (x × x)/(0.400 -x) = x^2/(0.400 - x) = 1.60 × 10^"-10}}$

0.400/(1.60 × 10^"-10") = 2.50 × 10^9

x ≪ 0.400

The equation reduces to

x^2/0.400 = 1.60 × 10^"-10"

x^2 = 0.400 × 1.60 × 10^"-10" = 6.40 × 10^"-11"

x = 8.00 × 10^"-6"

["OH"^"-"] = 8.00 × 10^"-6"color(white)(l) "mol/L"

"pOH" = -log["OH"^"-"] = -log(8.00 × 10^"-6") = 5.10

$\text{pH" = "14.00 - pOH" = "14.00 - 5.10} = 8.90$