Question

Calculate the volume (in mL) of 0.170 M NaOH that must be added to 345 mL of 0.0499 M 3-(N-Morpholino)propanesulfonic acid (MOPS) to give the solution a pH of 7.55. The pKa of MOPS = 7.18.

Answer 1

pH = pKa + log [MOPS-]/[HMOPS]

7.55 = 7.18 + log [MOPS-]/[HMOPS]

log [MOPS-]/[HMOPS] = 0.37

[MOPS-]/[HMOPS] = 2.34 ------------------------ eq.1

Rearranging this gives [MOPS-] = 2.34 [HMOPS]

Now, we know that [MOPS-] + [HMOPS] = 0.0499 M. ------------------- eq.2

Substituting the eq.1 in eq. 2 gives:
2.34 [HMOPS] + [HMOPS] = 0.0499

3.34 [HMOPS] = 0.0499

[HMOPS] = 0.0149 M

and [MOPS-] = 2.34 [HMOPS] = 2.34 * 0.0149 M = 0.0349 M

Now, since we have only 0.345 L of the solution, the moles of the basic form (which will equal the moles of NaOH added to the solution) is 0.345 L X 0.0349 mol/L = 0.0120 mol MOPS- = 0.0120 mol NaOH added

M = n/V ===> V = n / M

So, volume NaOH = 0.0120 mol / 0.170 mol/L = 0.0705 L NaOH = 70.5 mL NaOH