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

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

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)