## Chemistry and Chemical Reactivity (9th Edition)

The hydronium ion concentration for that $NH_4Cl$ solution is equal to $1.1 \times 10^{-5}M$. And the pH is equal to 4.98.
Analyze the ions of the salt: $Cl^-$: Conjugate base of $HCl$ (Strong acid). Therefore, it will not affect the pH of the water. $NH_4^+$: Conjugate acid of a weak base. So it is a weak acid. Calculate the pH for a $0.20M$ $NH_4^+$ solution. 1. Since $NH_4^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.56\times 10^{- 10}$ 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 the same for $[H_3O^+]$ is $\approx$ 0 M -$[H_3O^+] = [NH_3] = 0 + x = x$ -$[NH_4^+] = [NH_4^+]_{initial} - x$ 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. $Ka = \frac{[H_3O^+][NH_3]}{ [NH_4^+]}$ $Ka = 5.56 \times 10^{- 10}= \frac{x * x}{ 0.20}$ $Ka = 5.56 \times 10^{- 10}= \frac{x^2}{ 0.20}$ $x^2 = 0.20 \times 5.56 \times 10^{-10}$ $x = \sqrt { 0.20 \times 5.56 \times 10^{-10}} = 1.054 \times 10^{-5}$ Percent dissociation: $\frac{ 1.05 \times 10^{- 5}}{ 0.2} \times 100\% = 0.00527\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [NH_3] = x = 1.054 \times 10^{- 5}M$ And, since 'x' has a very small value (compared to the initial concentration): $[NH_4^+] \approx 0.20M$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.054 \times 10^{- 5})$ $pH = 4.98$