Answer to Question #112961 in General Chemistry for Nicolas

Question

If there are 0.045 mols of NH3 in 23.6 mL of solution, what is its concentration? Also,
what is the pH?

Answer 1

n=0.045 mols;

V = 23.6 mL= 0.0236 L;

Concentration C=n/V;

C=0.045/0.0236= 1.906 M;

Ammonia acts as a weak base in aqueous solution.

NH₃ + H₂O <=> NH₄⁺ + OH^-

The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant, K_b.

K_b=([NH₄⁺]×[OH^-])/[NH₃]

Ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take x M to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain

[NH₄⁺] = [OH^-] = x M;

[NH₃] = (1.906 − x) M;

K_b=(x×x)/(1.906 − x);

The dissociation constant for ammonia at 25 °C is 1.8×10^-5

1.8×10^-5 = x²/(1.906 − x);

Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation

1.906 − x ≈ 1.906

because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left.

1.8×10^-5 = x²/1.906;

x² = 3.4308×10^-5;

x = 0.005857;

[OH^-] = 0.005857 M;

pH + pOH = 14;

pOH = −log([OH^-]);

pH = 14 + log([OH^-]);

pH = 14 + log(0.005857) = 14 − 2.23 = 11.77;

Answer: 1.906 M; pH = 11.77