Question

In standardizing a solution of NaOH against 1.431 grams of KHP, the analyst uses 35.50 ml of the alkali and has to run back with 8.25 ml of acid (1 ml = 10.75 mg NaOH). What is the molarity of NaOH? 0.2598

Answer 1

Given the mass of KHP(C8H5KO4) taken = 1.431 g

Molecular mass of KHP = 204.2 g/mol

Hence moles of KHP = mass / molecular mass = 1.431 g / 204.2 g/mol = 0.00701 mol KHP

Let the molarity of NaOH be 'M'

Volume of NaOH taken, V = 35.50 mL = 0.03550 L

Hence initial moles of NaOH taken = MxV = 0.03550M mol

Now the excess of NaOH remained was titrated back with  8.25 ml of acid (1 ml = 10.75 mg NaOH).

Molecular mass of NaOH = 40.0 g/mol

Hence moles of excess NaOH remained = moles of NaOH titrated back to acid

= (10.75 mg NaOH / 1mL acid) x (1 mg mol NaOH / 40.0 mg NaOH) x (8.25 mL acid)

= 2.22 mg mol NaOH

= 2.22 x 10^-3 mol NaOH

Hence moles of NaOH titrated with KHP = 0.03550M mol - 2.22 x 10^-3 mol

The neutralization reaction of NaOH with KHP(C8H5KO4) is

NaOH + C8H5KO4 ------ > NaC8H4KO4 + H2O

1 mol, 1 mol -----------------1 mol ------------ 1 mol

In the above balanced reaction 1 mole of NaOH reacts with 1 mole of C8H5KO4.

Hence moles of NaOH titrated with KHP = Moles of KHP(C8H5KO4)

=>  0.03550M mol - 2.22 x 10^-3 mol = 0.00701 mol

=> 0.03550M mol = 0.00701 mol + 2.22 x 10^-3 mol = 0.00923 mol

=> M = 0.00923 /  0.03550 = 0.2598 M (answer)