Answer to Question #243474 in Chemistry for deevika

Question

During the course of an experiment, a student prepared a solution of sodium hydroxide by dissolving 3g of the salt in 100mL of water. Calculate the pH of the solution.

Answer 1

Solution:

sodium hydroxide = NaOH

The molar mass of NaOH is 40 g/mol

Hence,

(3 g NaOH) × (1 mol NaOH / 40 g NaOH) = 0.075 mol NaOH

n(NaOH) = 0.075 mol

Calculate the molarity of NaOH:

V(solution) = V(water) = 100 mL = 0.1 L

(0.075 mol NaOH) / (0.1 L) = 0.75 mol/L = 0.75 M NaOH

C(NaOH) = 0.75 M

NaOH → Na⁺ + OH^-

NaOH is a strong base.

Hence,

[OH^-] = [Na⁺] = C(NaOH) = 0.75 M

We can convert between [OH^-] and pOH using the following equation:

pOH = - log[OH^-]

pOH = - log(0.75) = 0.125

pOH = 0.125

 

For any aqueous solution at 25^∘C:

pH + pOH = 14

pH = 14 - pOH = 14 - 0.125 = 13.875 = 13.88

pH = 13.88

Answer: pH = 13.88