Question

1. What is the pH of a solution that has a hydroxide concentration of 1.9 X 10-5 M?

2. Lye has 400 mg of NaOH per teaspoon. Calculate the mass of battery acid that can be neutralized by 1 teaspoon of lye. Assume that battery acid is only 5% H2SO4 by mass?

Answer 1

1.

The hydroxide ion concentration of the solution is

JOH-] = 1.9 x 10-5 M

pH is defined as the negative logarithm of H+ ion concentration . i.e. .

pH = -log[H+

We know that the H+ concentration and OH- concentration in an aqueous solution are related as follows:

Ky = [+][OH-]=1.0 10-14

Hence, for our solution H+ concentration can be calculated as

[+][OH-] = 1.0 x 10-14 1.0 x 10-14 = [H+]= 1.9 x 10- 5 5 .26 x 10-10

Hence, the pH of the solution can be calculated as

pH = -log[H+1= -log(5.26 x 10-10) = 9.279 9.3

Hence, the pH of the solution is 9.3 approximately ( rounded to 2 significant figures).

2.

Assuming complete deprotonation of the acid , NaOH neutralises battery acid (H2SO4) as follows:

2NaOH(aq) + H2SO4(aq) + Na2SO4(aq) + 2H,00

Hence, 2 moles of NaOH neutralises 1 mol of H2SO4.

Amount of NaOH in the 1 tsp lye taken = 400 mg = 0.400 g

Molar mass of NaOH = 40.0 g/mol

Hence, number of moles of NaOH taken =

mass molar mass 0.400 g = 0.0100 mol 40.0 g/mol O

Hence, the number of moles of H2SO4 that can be neutralised by 0.0100 mol of NaOH is

1 mol H2SO4 x 0.0100 mol NaOH = 0.0050 mol H2SO4 2 mol NaOH

Molar mass of H2SO4 = 98.079 g/mol

Hence, the mass of H2SO4 that can be neutralised is

98.079 g - X 0.0050 mol H2SO4 0.4904 g 1 mol H.S04

The battery acid contains 5% H2SO4 by mass

Hence, there are 5 g of H2SO4 in 100 g battery acid.

Hence, the amount of battery acid that has 0.4904 g of H2SO4 is

100 g battery acid - x 0.4904 g 5g H2SO4 9.8 g battery acid

Hence, the mass of battery acid that will be neutralised by 1 tsp of lye is about 9.8 g. (rounded to 2 significant figures).