A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of NH₄⁺ is 0.20 M. If the equilibrium constant, Kb for NH₃ equals 1.8 × 10⁻⁵, what is the pH of this solution?

Question

Dr. Abubacker · Accepted Answer

[Solution from source] pKb = −log(1.8 × 10⁻⁵) = 4.74. pOH = pKb + log([NH₄⁺]/[NH₃]) = 4.74 + log(0.20/0.30) = 4.74 − 0.18 = 4.57, so pH = 14 − 4.57 = 9.43. Swapping the ratio gives pOH 4.92 and pH 9.08, the trap. Answer (3).

Q. A buffer solution is prepared in which the concentration of NH₃ is 0.30 M and the concentration of NH₄⁺ is 0.20 M. If the equilibrium constant, Kb for NH₃ equals 1.8 × 10⁻⁵, what is the pH of this solution?

Explanation

Reviewed by Dr. Abubacker

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