Question

# 1. A 0.312 g of an unknown acid (monoprotic) was dissolved in 26.5 mL of water...

1. A 0.312 g of an unknown acid (monoprotic) was dissolved in 26.5 mL of water and titrated with 0.0850 M NaOH. The acid required 28.5 mL of base to reach the equivalence point.

What is the molar mass of the acid? After 16.0 mL of base had been added in the titration, the pH was found to be 6.45. What is the Ka for the unknown acid?

Number of moles of Base = 28.5 mL * 0.085 M = 2.4225 mmoles

number of moles of acid = 2.4225 mmoles

because the acid is mono protic acid which requires same number of moles of the of base at equivalence point.

molar mass of the acid = 0.312g/2.4225 mmol = 128.8 g/mol

concentrtation of acid = 0.312/128.8 * 1000/26.5

Concentration of the acid = 0.09141 M

Number of moles of Base = 0.0850 * 16 = 1.36 mmol

Number of moles of Acid = 0.09141 * 26.5 = 2.4225 mmol

number of moles remaining = 2.4225 - 1.36 = 1.0625 mmol

Concentration of Acid = 1.0625/(28.5+16) = 0.02388 M

since it is weak acid,

pH = 1/2(pKa - log C)

6.45 = 1/2(pKa - log(0.02388))

pKa = 11.278

Ka = 10^-11.278

Ka = 5.2723 * 10^-12

#### Earn Coins

Coins can be redeemed for fabulous gifts.