Question

Calculate the pH of 20.00 mL of 0.150 M of a weak acid after 15.00 mL of 0.200 M NaOH is added to the solution. The Ka of the weak acid is 1.80 X 10-5. Write the answer in 3 significant figures. g

Answer 1

Answer:

pH = 8.84

Explanation:

The reaction of HA + NaOH is:

HA + NaOH → H₂O + Na⁺ + A⁻

20.00mL of 0.150M HA are:

0.020L × (0.150mol / L) = 3.00x10⁻³ moles of HA

Moles of NaOH are:

0.015L × (0.150mol / L) = 3.00x10⁻³ moles of NaOH

That means all HA reacts producing 3.00x10⁻³ moles A⁻. In 35mL of total volume: 3.00x10⁻³ moles A⁻ / 0.035L = 0.0857M.

The equilibrium of A⁻ in water is:

A⁻ + H₂O ⇄ HA + OH⁻

Where Kb is (Kw / Ka =  1.00x10⁻¹⁴ / 1.80x10⁻⁵ = 5.56x10⁻¹⁰) defined as:

Kb = [HA] [OH⁻] / [A⁻]

Equilibrium concentrations are:

[A⁻] = 0.0857M - X

[HA] = X

[OH⁻] = X

Replacing in Kb formula:

5.56x10⁻¹⁰ = [X] [X] / [0.0857M - X]

4.77x10⁻¹¹ - 5.56x10⁻¹⁰X = X²

0 = X² + 5.56x10⁻¹⁰X - 4.77x10⁻¹¹

Solving for x:

X = -6.907x10⁻⁶ M  → False answer → There is no negative concentrations

X = 6.906x10⁻⁶ M → Right answer

As X = [OH⁻] and pOH = -log [OH⁻], pOH =-log 6.906x10⁻⁶ M = 5.16

As 14 = pH + pOH

pH = 14 - 5.16

pH = 8.84

Answer 2

Answer:

pH of the weak acid is = 4.74 to 3 sig. fig.

Explanation:

Firstly, find the number of moles, n , in the reaction

c = n/V  where c = concen. and V = volume

$n_{acid}$ = 0.15 M × 20 ×$10^{-3}$ = 0.003

$n_{base}$ =  0.2 × 15 × $10^{-3}$ = 0.003

mole ratio = 1 : 1 ⇒ $\frac{[A^{-} ]}{[HA]}$ = 1

But pH = pKa + log $\frac{[A^{-} ]}{[HA]}$

∴ pH = -log (1.8×$10^{-5}$)

pH = 4.74 to 3 sig. fig.