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Question

Chemistry Question on Equilibrium

What is the pHpH of a 0.5M0.5\, M aqueous solution of NH4ClNH_4Cl with 0.2MNH30.2\, M \,NH_3? (pKa=9.25) (pK_a = 9.25)

A

8.858.85

B

10.1510.15

C

5.155.15

D

4.354.35

Answer

8.858.85

Explanation

Solution

NH4+NH_{4}^{+} and NH3NH_3 are conjugate acid-base pair. For conjugate add-base pair pKa+pKb=14 \to pK_a + pK_b = 14 pKb=149.25=4.75pK_b = 14 - 9.25 = 4.75 pOH=pKb+log[Conjugate acid NH4+][Weak base NH3]pOH = pK_b + log \frac{[{\text{Conjugate acid }} NH_{4}^{+}]}{[\text{Weak base } NH_3]} =4.75+log0.50.2 = 4.75 + log \frac{0.5}{0.2} =4.75+0.40=5.15= 4.75 + 0.40 = 5.15 pH=14pOH=145.15=8.85pH = 14 - pOH = 14 - 5.15 = 8.85.