## Chemistry (4th Edition)

$pH = 5.39$
$C_2H_5NH_3I$: $I^-$ does not act as an electrolyte, but the cation of this salt is an acid in water: 1. Since $C_2H_5NH_3^+$ is the conjugate acid of $C_2H_5NH_2$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $5.6\times 10^{- 4} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 5.6\times 10^{- 4}}$ $K_a = 1.8\times 10^{- 11}$ 2. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium: For the reaction: $C_2H_5NH_2(aq) + H_2O(l) \lt -- \gt C_2H_5NH_3^+(aq) + H_3O^+(aq)$ Remember: Reactants get a '-x', and products get a '+x' The initial concentration of $C_2H_5NH_3^+$ is : 0 M, and $[OH^-]$ is $\approx$ 0 M -$[OH^-] = [C_2H_5NH_3^+] = 0 + x = x$ -$[C_2H_5NH_2] = [C_2H_5NH_2]_{initial} - x$ For approximation, we are going to consider $[C_2H_5NH_2]_{initial} = [C_2H_5NH_2]$ 3. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_2H_5NH_2]}{ [C_2H_5NH_3^+]}$ $Ka = 1.8 \times 10^{- 11}= \frac{x * x}{ 0.91}$ $Ka = 1.8 \times 10^{- 11}= \frac{x^2}{ 0.91}$ $1.6 \times 10^{- 11} = x^2$ $x = 4 \times 10^{- 6}$ Percent dissociation: $\frac{ 4 \times 10^{- 6}}{ 0.91} \times 100\% = 0.00044\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [C_2H_5NH_2] = x = 4 \times 10^{- 6}M$ And, since 'x' has a very small value (compared to the initial concentration): $[C_2H_5NH_3^+] \approx 0.91M$ 4. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 4 \times 10^{- 6})$ $pH = 5.39$