Question

A buffer solution is made up of 100 mL 1.0696 M acetic acid, and 100 mL 1.0410 M sodium acetate.

A) Calculate the pH of 75 mL of the buffer solution after the addition of 1 mL of 3 M HCl.

B) Calculate the pH of 75 mL of the buffer solution after the addition of 10 mL of 3M HCl.

Answer 1

Number of moles of acetic acid = 1.0696 M * 0.1 L = 0.10696 mol

Number of moles of acetate = 1.0410 M * 0.1 L = 0.1041 mol

Total volume of the solution = 100 + 100 = 200 mL = 0.2 L

Conc. of acetic acid in buffer = moles/total volume = 0.10696 mol/0.2 L = 0.5348 M

Conc. of acetic acid in buffer = moles/total volume = 0.1041 mol/0.2 L = 0.5205 M

Number of moles of acetic acid in 75 mL of buffer = 0.5348 M * 0.075 L = 0.04011 mol

Number of moles of acetate in 75 mL of buffer = 0.5205 M * 0.075 L = 0.039038 mol

Number of moles of HCl added = 1 mL * 3 M = 3 mmol = 0.003 mol

When HCl adds to the buffer solution , it reacts with CH3COO– to form CH3OOH. So the conc. of CH3COO– decreases and conc. of CH3COOH increases.

	base		acid
	CH3COO–	H+ ------>	CH3COOH
Initial	0.039038		0.04011
Add	0	0.003	0
change	-0.003	-0.003	0.003
Equilibrium	0.036038		0.04311

pH = pKa + log[base]/[acid]

pH = – logKa +   log[CH3COO–]/[CH3COOH]

pH = – log(1.8*10^-5) +   log[0.036038]/[0.04311] = 4.745 – 0.07782 = 4.67

========================================================================================================

Number of moles of acetic acid = 1.0696 M * 0.1 L = 0.10696 mol

Number of moles of acetate = 1.0410 M * 0.1 L = 0.1041 mol

Total volume of the solution = 100 + 100 = 200 mL = 0.2 L

Conc. of acetic acid in buffer = moles/total volume = 0.10696 mol/0.2 L = 0.5348 M

Conc. of acetic acid in buffer = moles/total volume = 0.1041 mol/0.2 L = 0.5205 M

Number of moles of acetic acid in 75 mL of buffer = 0.5348 M * 0.075 L = 0.04011 mol

Number of moles of acetate in 75 mL of buffer = 0.5205 M * 0.075 L = 0.039038 mol

Number of moles of HCl added = 10 mL * 3 M = 30 mmol = 0.03 mol

When HCl adds to the buffer solution , it reacts with CH3COO– to form CH3OOH. So the conc. of CH3COO– decreases and conc. of CH3COOH increases.

	base		acid
	CH3COO–	H+ ------>	CH3COOH
Initial	0.039038		0.04011
Add	0	0.03	0
change	-0.03	-0.03	0.03
Equilibrium	0.009038		0.07011

pH = pKa + log[base]/[acid]

pH = – logKa +   log[CH3COO–]/[CH3COOH]

pH = – log(1.8*10^-5) +   log[0.009038]/[0.07011] = 4.745 – 0.88973 = 3.85

========================================================================================================