# Chapter 16 - Questions and Problems - Page 774: 16.103

$pH = 4.82$

#### Work Step by Step

$NH_4Cl$: $Cl^-$ is not an electrolyte, but $NH_4^+$ act as an acid in water. 1. Since $NH_4^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $NH_4^+(aq) + H_2O(l) \lt -- \gt NH_3(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $NH_3$ is : 0 M, and $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [NH_3] = 0 + x = x$ -$[NH_4^+] = [NH_4^+]_{initial} - x$ For approximation, we are going to consider $[NH_4^+]_{initial} = [NH_4^+]$ 3. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][NH_4^+]}{ [NH_3]}$ $Ka = 5.6 \times 10^{- 10}= \frac{x * x}{ 0.42}$ $Ka = 5.6 \times 10^{- 10}= \frac{x^2}{ 0.42}$ $2.4 \times 10^{- 10} = x^2$ $x = 1.5 \times 10^{- 5}$ Percent dissociation: $\frac{ 1.5 \times 10^{- 5}}{ 0.42} \times 100\% = 0.0037\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [NH_4^+] = x = 1.5 \times 10^{- 5}M$ And, since 'x' has a very small value (compared to the initial concentration): $[NH_3] \approx 0.42M$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.5 \times 10^{- 5})$ $pH = 4.82$

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