# Chapter 17 - Questions and Problems - Page 824: 17.9

The pH of this buffer is equal to $8.89$

#### Work Step by Step

1. Since $NH_4$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ 2. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.556 \times 10^{- 10})$ $pKa = 9.255$ 3. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.15}{0.35}$ - 0.4286: It is. 4. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.15}{5.556 \times 10^{-10}} = 2.7\times 10^{8}$ - $\frac{0.35}{5.556 \times 10^{-10}} = 6.3\times 10^{8}$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.255 + log(\frac{0.15}{0.35})$ $pH = 9.255 + log(0.4286)$ $pH = 9.255 + ( -0.3679)$ $pH = 8.887$

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