Question

MOPS – 3(N-morpholino)propanesulfonic acid (structure below) is a common buffer used in biochemistry and biology. Consider a 0.100 M MOPS/Na+•(MOPS-) buffer solution (200.0 mL). pKa (MOPS) = 7.20 (0.1 M MOPS AND 0.1 M Na+ MOPS-)

a. Calculate the pH of the buffer.

b. Calculate the pH when 1.0 mL of 1.0 M NaOH is added to the buffer.

c. Calculate the pH when 5.0 mL of 1.0 M NaOH is added to the buffer.

Answer 1

Given: Concentration of buffer solution = 0.100 M in 200.0 mL

pK_a (MOPS) = 7.20

a) Henderson-Hasselbatch equation to calculate pH of the buffer solution is given by

$pH = pK_{a} + log\frac{[salt]}{[acid]}$

$pH = 7.20 + log\frac{[0.1]}{[0.1]}$

pH = 7.20

b) MOPS + NaOH -->  Na⁺MOPS^- + H₂O

Buffer Initially contains,

MOPS = 0.100 M x 200 mL = 0.002 moles

Na⁺MOPS^- = 0.100 M x 200 mL = 0.002 moles

Addition of 1.0mL of 1.0M NaOH to the buffer

OH^- = 0.001L x 1.0mol/L = 0.001 moles

MOPS decreased by 0.001 moles and Na⁺MOPS^- increased by 0.001 moles

[MOPS] = 0.002 -0.001 = 0.001 moles

[Na⁺MOPS^-] = 0.002 + 0.001 = 0.003 moles

$pH = 7.20 + log\frac{[0.003]}{[0.001]}$

pH = 7.20 + log 3

= 7.20 + 0.477

= 7.677

c) MOPS + NaOH -->  Na⁺MOPS^- + H₂O

Buffer Initially contains,

MOPS = 0.100 M x 200 mL = 0.002 moles

Na⁺MOPS^- = 0.100 M x 200 mL = 0.002 moles

Addition of 5.0mL of 1.0M NaOH to the buffer

OH^- = 0.005L x 1.0mol/L = 0.005 moles

MOPS decreased by 0.005 moles and Na⁺MOPS^- increased by 0.005 moles

[MOPS] = 0.002 -0.005 = -0.003 moles

[Na⁺MOPS^-] = 0.002 + 0.005 = 0.007 moles

$pH = 7.20 + log\frac{[0.007]}{[-0.003]}$

pH = 7.20 - log 2.33

= 7.20 - 0.367

= 6.833

Answer 2

Formula is right. He converted wrong. 200ml is 0.2L. Just adjust accordingly.