A buffer solution contains 0.486 M NaHCO3 and 0.345 M Na2CO3. If 0.0321 moles of sodium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution

pH = 10.51

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

0,150L of these substances are:

0,150L × 0,486M NaHCO₃ = 0.0729mol NaHCO₃

0,150L × 0,345M Na₂CO₃ = 0.05175mol Na₂CO₃

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.0321 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.05175mol + 0.0321mol: 0.08385mol

HCO₃⁻: 0.0729mol - 0.0321 mol: 0.0408mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.08385mol / 0.0408mol

pH = 10.51

I hope it helps!

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