Question

You have a 1.153 g sample of an unknown solid acid, HA, dissolved in enough water to make 20.00 mL of solution. HA reacts with KOH(aq) according to the following balanced chemical equation: HA(aq)+KOH(aq)--->KA(aq)+H2O(l)

if 13.40 mL of 0.715 M KOH is required to titrate the unknown acid to the equivalence point, what is the concentration of the unknown acid? ALSO What is the molar mass of HA?

Answer 1

HA(aq)+KOH(aq)--->KA(aq)+H2O(l)

1 mole   1 mole

HA                                                                        KOH

M1 =                                                                 M2   = 0.715M

V1   = 20ml                                                        V2   = 13.4ml

n1 = 1                                                                n2 = 1

                 M1V1/n1     =    M2V2/n2

                   M1            = M2V2n1/V1n2

                                   = 0.715*13.4*1/20*1

                                   = 0.48M

the concentration of HA   = 0.48M

HA(aq)+KOH(aq)--->KA(aq)+H2O(l)

no of moles of KOH   = molarity * volume in L

                                  = 0.715*0.0134    = 0.00958moles

from balanced equation

1 mole of KOH react with 1 mole of HA

0.00958 moles of KOH react with 0.00958 moles of HA

molar mass of HA   = mass of HA/no of moles of HA

                               = 1.153/0.00958    = 120.35g/mole

molar mass of acid HA is 120.35g/mole