Question

A 5.00mL aliquot of H2SO4 is neutralized with 23.55mL of a 0.1000M NaOH solution.

a. calculate the pH of the acid.

b. calculate the % (w/v) of the acid.

Answer 1

We know from the law of acidimetry and alkalimetry,

V₁S₁ = V₂S₂ where, V₁= volume of H2SO4 solution = 5 mL

S₁ = strength of H2SO4 solution
V₂=  volume of NaOH solution = 23.55 mL

S₂= strength of NaOH solution =0.1M

Therefore,  strength of H2SO4 solution ( S₁ ) =  V₂S₂ / V₁ = 23.55 x 0.1 / 5 (M) = 0.471 (M)

(a) Then pH of H2SO4 solution = - log [H+] = - log [ 2 x 0.471 ] = - log 0.942 = 0.0259 (ANSWER)

(b) 1mole H2SO4 = 98.079 g

So, 0.471 mole  H2SO4 = 98.079 x 0.471 = 46.195 g

Then 1000 mL of H2SO4 contain 46.195 g of acid

Therefore, 100 mL of  H2SO4 contain { (46.195 x 100 ) /1000 } g = 4.6195 g of acid

So, % (w/v) of the acid is 4.6195 g /100 mL (ANSWER)