Question

You have an aqueous solution with the following bases, what is the pH of the mixture?
[NaOH] = 100 mL of solution containing 25 g of 98% pure NaOH, pKa = ----
[NH3] = 0.05 M, pKa = 4.8
[NaF] = 0.01 M, pKa = 10.8
[KOH] = 0.1 L of solution containing 50 g of KOH, pKa = ----

Answer 1

1) The supplied NaOH is 98% pure; therefore, mass of pure NaOH in 25 g sample = 98%*(25 g) = (98/100)*(25 g) = 24.50 g.

Molar mass of NaOH = (1*22.989 + 1*15.999 + 1 *1.008) g/mol = 39.996 g/mol.

Mole(s) of NaOH corresponding to 24.50 g = (24.50 g)/(39.996 g/mol) = 0.6126 mole.

The NaOH is dissolved in 100 mL water; therefore,

[NaOH] = [OH^-] = (0.6126 mole)/[(100 mL)*(1 L)/(1000 mL)]

= 6.126 mol/L

= 6.126 M ([..] denote concentrations in M).

We define pOH = -log [OH^-]

= -log (6.126 M)

= 0.7872

We know that

pH + pOH = 14

=====> pH = 14 – pOH

=====> pH = 14 – 0.7872

=====> pH = 13.2128

=====> pH = 13.2 (ans).

NaOH is a strong base and ionizes completely; hence, the concept of pK_a doesn’t apply.

2) pK_a = 4.8.

Since NH₃ is a base, we will work with pK_b. We know that

pK_a + pK_b = 14

Therefore,

pK_b = 14 – pK_a

= 14 – 4.8

= 9.2

=====> K_b = antilog (-pK_b) [We define pK_b as pK_b = -log K_b.]

=====> K_b = antilog (-9.2)

=====> K_b = 6.31*10^-9

NH₃ ionizes as

NH₃ (aq) + H₂O (l) <======> NH₄⁺ (aq) + OH^- (aq)

K_b = [NH₄⁺][OH^-]/[NH₃]

=====> 6.31*10^-9 = (x)(x)/(0.05 – x)

=====> 6.31*10^-9 = x²/(0.05 – x)

Since K_b is small, assume x << 0.05 M and thus, (0.05 – x) M ≈ 0.05 M; therefore,

6.31*10^-9 = x²/(0.05)

=====> x² = 6.31*10^-9*0.05

=====> x = 3.155*10^-10

=====> x = 1.7762*10^-5

Thus, [OH^-] = 1.7762*10^-5 M and

pOH = -log [OH^-]

= -log (1.7762*10^-5 M)

= 4.7506

We know that

pH + pOH = 14

=====> pH = 14 – pOH

=====> pH = 14 – 4.7506

=====> pH = 9.2494

=====> pH = 9.25 (ans).

3) pK_a = 10.8.

Since F^- is a base, we will work with pK_b. We know that

pK_a + pK_b = 14

Therefore,

pK_b = 14 – pK_a

= 14 – 10.8

= 3.2

=====> K_b = antilog (-pK_b) [We define pK_b as pK_b = -log K_b.]

=====> K_b = antilog (-3.2)

=====> K_b = 6.31*10^-4

F^- ionizes as

F^- (aq) + H₂O (l) <======> HF (aq) + OH^- (aq)

K_b = [HF][OH^-]/[F^-]

=====> 6.31*10^-4 = (x)(x)/(0.01 – x)

=====> 6.31*10^-4 = x²/(0.01 – x)

Since K_b is small, assume x << 0.01 M and thus, (0.01 – x) M ≈ 0.01 M; therefore,

6.31*10^-4 = x²/(0.01)

=====> x² = 6.31*10^-4*0.01

=====> x = 6.31*10^-6

=====> x = 2.5120*10^-3

Thus, [OH^-] = 2.5120*10^-3 M and

pOH = -log [OH^-]

= -log (2.5120*10^-3 M)

≈ 2.600

We know that

pH + pOH = 14

=====> pH = 14 – pOH

=====> pH = 14 – 2.600

=====> pH = 11.40

=====> pH = 11.4 (ans).

4) Molar mass of KOH = (1*39.098 + 1*15.999 + 1 *1.008) g/mol = 56.987 g/mol.

Mole(s) of KOH corresponding to 50 g = (50 g)/(56.987 g/mol) = 0.8774 mole.

The NaOH is dissolved in 0.1 L water; therefore,

[KOH] = [OH^-] = (0.8774 mole)/(0.1 L)

= 8.774 mol/L

= 8.774 M ([..] denote concentrations in M).

We define pOH = -log [OH^-]

= -log (8.774 M)

= 0.9432

We know that

pH + pOH = 14

=====> pH = 14 – pOH

=====> pH = 14 – 0.9432

=====> pH = 13.0568

=====> pH = 13.05 (ans).

KOH is a strong base and ionizes completely; hence, the concept of pK_a doesn’t apply.