Question

A 5.65 −g sample of a weak acid with Ka=1.3×10−4 was combined with 5.20 mL of 5.90 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.30. What is the molar mass of the weak acid?

**please show work

Answer 1

Answer – We are given, mass of weak acid = 6.56 g , Ka = 1.3*10^-4

[NaOH] = 5.90 M , volume = 5.20 mL

Total volume = 750 mL , pH = 4.30

We assume HA is the weak acid

Moles of NaOH = 0.00520 L * 5.90 M = 0.0307 moles

The pH is acidic means all NaOH gets consumed

HA + NaOH -----> H₂O + A^-

           0.0307                0.0307

[A^-] = 0.0307 moles / 0.750 L = 0.0409 M

First we need to calculate the pKa from Ka

We know formula

pKa = - log Ka

        = - log 1.3*10^-4

        = 3.89

We know Henderson–Hasselbalch equation –

pH = pKa + log [A^-] /[HA]

so, log 0.0409 /[HA] = pH – pKa

                              = 4.30-3.89

                              = 0.414

So taking antilog from both side

0.0409 /[HA] = 2.60

So, [HA] = 0.0409 M / 2.60

                = 0.0158 M

S, moles of HA = 0.0158 M * 0.750 L

                          = 0.0118 moles

So molar mass of HA = 5.65 g / 0.0118 mole

                                   = 477.67 g/moles