Question

You are asked to prepare a pH=4.00 buffer starting from 1.50 L of 0.0200M solution of benzoic acid (C6H5COOH) and an excess of sodium benzoate (C6H5COONa).

Part A

What is the pH of the benzoic acid solution prior to adding sodium benzoate?

Express your answer using three significant figures.

Part B

How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

Express your answer using two significant figures.

Answer 1

pH= 4

V= 1.5 L

M = 0.02 benzoic Acid

+ salt sodium benzoate

A)

pH prior adding sodium benzoate....

From database: Ka= 6.5 x 10-5

Benzoic acid dissociates in water as

C6H5COOH ? H+ + C6H5COO-

The formula for Ka is

Ka = [H+][B-]/[HB]

where
[H+] = concentration of H+ ions
[B-] = concentration of conjugate base ions
[HB] = concentration of undissociated acid molecules
for a reaction HB ? H+ + B-

Benzoic acid dissociates one H+ ion for every

C6H5COO- ion, so [H+] = [C6H5COO-].

Let x represent the concentration of H+ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the Ka equation

Ka = x � x / (C -x)
Ka = x�/(C - x)
(C - x)Ka = x�
x� = CKa - xKa
x� + Kax - CKa = 0

x = [-Ka + (Ka� + 4CKa)^0.5]/2

C= 0.02, ka = 6.5 x 10-5

x = 0.00117

there are two solutions for x.

Since x represents a concentration of ions in solution, the value for x cannot be negative.

Now, having [H+]

get ph with

pH = -log[H+] = -log ( 0.00117) = 2.93

pH= 2.93

Part B

grams of NaB to prepare the buffer...

pH = pKa + log [ Salt ] / [ Acid ]

4 = -log(kA) + log (salt/acid)

calculate pKa

pKa = - log(Ka) = -log (6.5*10^-5= 4.18

substitute all in equation

4 = 4.18 + log ([salt]/0.02)

-0.18 = log ([salt]/0.02)

10^-0.18 = ([salt]/0.02)

0.66*.02 = [salt]

0.99 = [sallt]

but we need mass not concnetration...

M = mole / Vol

M*vol = mole ... then we will change this to mass

0.99*1.5L = mole (assume no volume change)

1.489 moles

MW of NaB = 144 g/gmol

mass = mole*MW = 1.489*144 g/gmol = 214 grams of NaB are needed