Answer:
ΔpH = 0.20
Explanation:
The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2
HCO₃⁻ ⇄ H⁺ + CO₃²⁻
There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.342mol / 0.479mol
<em>pH = 10.05</em>
NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:
NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺
0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:
CO₃²⁻: 0.342mol + 0.091mol: 0.433mol
HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol
Using Henderson-Hasselbalch formula:
pH = pka + log [Base] / [Acid]
pH = 10.2 + log 0.433mol / 0.388mol
pH = 10.25
That means change of pH, ΔpH is:
ΔpH = 10.25 - 10.05 = <em>0.20</em>
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I hope it helps!
