Question

Calculate the pH of a 1L aqueous solution containing:

a) 20mL of 4M HCl

b) 100mL of 2M NaOH

c) 50mL of 100mM acetic acid and 200mL of 150mM potassium acetate (pKa is 4.76)

Please show work, thanks!!

Answer 1

a)      pH of a 1L aqueous solution containing 20mL of 4M HCl

M₁ = 4 M , V₁ = 20 mL = 0.020 L

M₂ = ?, V₂ = 1 L

M₁V₁ = M₂V₂

M₂ = (4 M x 0.020L)/1 L = 0.08 M

Concentration of 1L aqueous solution is 0.08 M.

Therefore the pH = -log(0.08) = - (-1.096) = 1.096

b)      pH of a 1L aqueous solution containing 100mL of 2M NaOH

M₁ = 2 M , V₁ = 100 mL = 0.100 L

M₂ = ?, V₂ = 1 L

M₁V₁ = M₂V₂

M₂ = (2 M x 0.100L)/1 L = 0.2 M

Concentration of 1L aqueous solution is 0.2 M.

Therefore the pOH = -log(0.2) = - (-0.698) = 0.698

pH = 14-0.698 = 13.302

c)      Henderson-Hasselebalch equation:

pH = pK_a + log ([A^-]/[HA])

[A^-] = molar concentration of a conjugate base

[HA] = molar concentration of a weak acid (M)

[HA] = 100 mM x 50 mL = (0.1 mol/L)x(0.050 L) = 0.005 moles

[A^-] = 150 mM x 200 mL = (0.15 mol/L)x(0.200 L) = 0.03 moles

Given that: pKa = 4.76

pH = 4.76 + log ([0.03]/[0.05]) = 4.76 – 0.221 = 4.539