Question

a) Calculate the volume (in ML) of 0.170 M NaOH that must be added to 365 mL of 0.0515 M 3-(N-Morpholino) propanesulfonic acid (MOPS) to give the solution a pH of 7.55. The pKa of MOPS =7.18.

b) A buffer with a pH of 4.24 contains 0.11 M of sodium benzoate and 0.10 M of benzoic acid. What is the concentration of [H+] in the solution after the addition of 0.056 mol of HCl to a final volume of 1.6 L? Assume that any concentration of HCl to the volume is negligible.

Answer 1

Answer – a) We are given, [MOPS] = 0.0515 M , volume = 365 mL = 0.365 L

pKa = 7.18, pH = 7.55

moles of MOPS = 0.0515 M * 0.365 L =0.0188 moles

We know Henderson Hasselbalch equation-

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

7.55 = 7.18 + log [MOPS^-] / 0.0188

7.55-7.18 = log [MOPS^-] / 0.0188

0.37= log [MOPS^-] / 0.0188

Taking antilog from both side

2.34 = MOPS^- / 0.0188

Moles of MOPS^- = 0.0188 * 2.34

                   = 0.0441 M

Moles of NaOH = moles of MOPS^-

So, moles of NaOH = 0.0441 moles

So, volume of NaOH = 0.0441 mole / 0.17 M

                               = 0.259 L

                               = 25.9 mL

b) We are given, [benzoate] = 0.11 M , [benzoic acid] = 0.10 M

volume = 1.6 L, moles of HCl = 0.056 moles , pKa = 4.19 ,

moles of benzoic acid = 0.10 *1.6 L = 0.16 moles

moles of benzoate = 0.11 M * 1.6 L = 0.176 moles

when we added the moles of HCl then there is moles of acid increase and moles of conjugate base decrease

New moles

moles of benzoic acid = 0.16 moles + 0.056 moles = 0.216

moles of benzoate = 0.176 moles - 0.056 moles =0.120

[benzoic acid] = 0.216 moles / 1.6 L = 0.135 M

[benzoate] = 0.120 moles / 1.6 L = 0.075 M

We know Henderson Hasselbalch equation-

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

      = 4.19 + log 0.075 / 0.135

    = 3.93

So, [H⁺] = 10^-pH

              = 1.16*10^-3 M