Question

A 5.079 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 26.00 mL of the resulting acid solution is then titrated with 0.09454 M NaOH. The pH after the addition of 18.00 mL of the base is 5.54, and the endpoint is reached after the addition of 47.60 mL of the base. Please see Titration to Determine Molecular Weight for assistance.

(a) How many moles of acid were present in the 26.00 mL sample? mol

(b) What is the molar mass of the acid? g/mol

(c) What is the pKa of the acid?

Answer 1

Answer – We are given, mass of sample of solid weak, monoprotic acid = 5.079 g. volume of acid used for the titration = 26.00 mL , [NaOH] = 0.09454 M,

Volume of NaOH used for titration = 47.60 mL , when volume of 18.0 mL of NaOH added then , pH = 5.54.

a) First we need to calculate the moles of base used for the titration

we know,

moles = molarity * volume (L)

           = 0.09454 M * 0.0476 L

           = 0.0045 moles

We know given acid is monoprotic and NaOH is also monobasic, so

HA + NaOH ----> NaA + H₂O

So, moles of NaOH = moles of HA = 0.0045 moles

b) We already calculated the moles of weak monoprotic acid and we know the mass of weak, monoprotic acid, so

molar mass of weak, monoprotic acid = mass / moles

                                                              = 5.079 g / 0.0045 moles

                                                              = 1128.6 g/mol

c) We know, pH after the addition of 18.00 mL of the base is 5.54

So, first we need to calculate the we added the moles of NaOH

moles of NaOH = 0.09454 M * 0.018 L

                       = 0.00170 moles

So, same number moles of HA also reacted with NaOH and form the A^-

So, moles of HA = 0.00170 moles , moles of A^- = 0.00170 moles

So, this is the half equivalence point and we know at half equivalence point

pH = pKa

so, pKa = 5.54.