Answer to Question #126065 in Chemistry for rushil

Question

What is the [H₂CO₃], [HCO₃^-] and pH of a buffer solution which is composed of 25 mL of 0.05 M H₂CO₃ and 475 mL of 0.263 M NaHCO₃? (K_a for H₂CO₃ = 4.2 x 10^-7).

Answer 1

Solution:

V(buffer solution) = V(H₂CO₃) + V(NaHCO₃) = 25 mL + 475 mL = 500 mL

[H₂CO₃] = C(H₂CO₃) × V(H₂CO₃) / V(solution) = (0.05 M) × (25 mL / 500 mL) = 0.0025 M

[H₂CO₃] = 0.0025 M

[HCO₃^-] = C(NaHCO₃) × V(NaHCO₃) / V(solution) = (0.263 M) × (475 mL / 500 mL) = 0.24985 M = 0.25 M

[HCO₃^-] = 0.2500 M

To calculate pH of a buffer solution Henderson-Hasselbalch Equation can be used.

Henderson-Hasselbalch Equation can be expressed as:

[H⁺] = K_a × ([HA]/[A^-])

take –log of both sides of equation

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

[A^-] = [HCO₃^-]

[HA] = [H₂CO₃]

pH = pK_a + log([HCO₃^-] / [H₂CO₃])

pK_a = -log(K_a) = -log(4.2×10^-7) = 6.376 = 6.38

Thus:

pH = 6.38 + log(0.2500 / 0.0025) = 6.38 + 2 = 8.38

pH = 8.38

Answer:

[H₂CO₃] = 0.0025 M

[HCO₃^-] = 0.2500 M

pH = 8.38