Question

Sodium carbonate (0.3379 g, FW: 105.99) was dissolved in deionized water to give a solution with a total volume of 250.0 mL. What was the pH of the resulting solution

Answer 1

pH = 11.179

Step-by-step explanation:

The sodium carbonate, Na2CO3, is in equilibrium with water as follows:

Na2CO3 + H2O ⇆ NaHCO3 + OH⁻

Where Kb is 1.786x10⁻⁴:

Kb = 1.786x10⁻⁴ = [NaHCO3] [OH-] / [Na2CO3]

Where [NaHCO3] = [OH-] = X, our incognite

And [Na2CO3] is molar concentration of sodium carbonate:

[Na2CO3] = 0.3379g * (1mol / 105.99g) = 3.188x10⁻³ moles Na2CO3

In 250.0mL = 0.2500L:

3.188x10⁻³ moles Na2CO3 / 0.2500L = 0.01275M

Replacing in Kb expression:

1.786x10⁻⁴ = [NaHCO3] [OH-] / [Na2CO3]

1.786x10⁻⁴ = [X] [X] / [0.01275]

2.278x10⁻⁶ = X²

X = [OH⁻] = 1.509x10⁻³M

As pOH = -log [OH⁻] = 2.821

And pH = 14 - pOH

pH = 11.179