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) + H20 (l)If 12.15 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?What is the molar mass of HA?

Answer 1

Answer:

The concentration of the unknown acid (HA) is 0.434M

The molar mass of HA is 13.3g/mole

Explanation:

DETERMINATION OF MOLARITY OF THE UNKNOWN ACID

CaVa/CbVb = Na/Nb

From the equation of reaction and at equivalence point, Na = Nb = 1

Therefore, CaVa = CbVb

Va (volume of acid solution) = 20mL = 20/1000 = 0.2L

Cb (concentration of KOH) = 0.715M

Vb (volume of KOH) = 12.15mL

Ca (concentration of acid) = CbVb/Va

Ca = 0.715M × 12.15mL/20mL = 0.434M

DETERMINATION OF MOLAR MASS OF HA

Number of moles of acid = concentration of acid × volume of acid solution in liters = 0.434 × 0.2 = 0.0868mole

Molar mass of HA = mass/number of moles = 1.153g/0.0868mole = 13.3g/mole