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A buffer solution is prepared in which the concentration of NH3NH_3 is 0.30 M and the concentration of NH4+NH_4^+ is 0.20 M. If the equilibrium constant, KbK_b for NH3NH_3 equals 1.8×10−51.8 \times 10^{-5}, then what is the pH of this solution? (log⁡1.8=0.25\log 1.8 = 0.25; log⁡0.67=−0.176\log 0.67 = -0.176)

A

9.43

B

11.72

C

8.73

D

9.08

Step-by-Step Solution

According to the Henderson-Hasselbalch equation for a basic buffer : pOH=pKb+log⁡[Conjugate acid][Base]pOH = pK_b + \log \frac{[\text{Conjugate acid}]}{[\text{Base}]} Given: Kb=1.8×10−5K_b = 1.8 \times 10^{-5} pKb=−log⁡(1.8×10−5)=5−log⁡1.8=5−0.25=4.75pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log 1.8 = 5 - 0.25 = 4.75 [Conjugate acid]=[NH4+]=0.20 M[\text{Conjugate acid}] = [NH_4^+] = 0.20 \text{ M} [Base]=[NH3]=0.30 M[\text{Base}] = [NH_3] = 0.30 \text{ M} Substitute the values into the equation: pOH=4.75+log⁡(0.200.30)=4.75+log⁡(0.67)pOH = 4.75 + \log \left(\frac{0.20}{0.30}\right) = 4.75 + \log(0.67) pOH=4.75−0.176=4.574pOH = 4.75 - 0.176 = 4.574 We know that at 298 K, pH+pOH=14pH + pOH = 14 . Therefore, pH=14−pOH=14−4.574=9.426≈9.43pH = 14 - pOH = 14 - 4.574 = 9.426 \approx 9.43

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