## Chemistry: Atoms First (2nd Edition)

$pH = 1.96$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_2H_2{ClO_2}^-] = x$ -$[HC_2H_2ClO_2] = [HC_2H_2ClO_2]_{initial} - x = 0.1 - x$ For approximation, we consider: $[HC_2H_2ClO_2] = 0.1M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_2H_2{ClO_2}^-]}{ [HC_2H_2ClO_2]}$ $Ka = 1.35 \times 10^{- 3}= \frac{x * x}{ 0.1}$ $Ka = 1.35 \times 10^{- 3}= \frac{x^2}{ 0.1}$ $1.35 \times 10^{- 4} = x^2$ $x = 1.162 \times 10^{- 2}$ Percent dissociation: $\frac{ 1.162 \times 10^{- 2}}{ 0.1} \times 100\% = 11.62\%$ %dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 1.35 \times 10^{- 3}= \frac{x^2}{ 0.1- x}$ $1.35 \times 10^{- 4} - 1.35 \times 10^{- 3}x = x^2$ $1.35 \times 10^{- 4} - 1.35 \times 10^{- 3}x - x^2 = 0$ $\Delta = (- 1.35 \times 10^{- 3})^2 - 4 * (-1) *( 1.35 \times 10^{- 4})$ $\Delta = 1.823 \times 10^{- 6} + 5.4 \times 10^{- 4} = 5.418 \times 10^{- 4}$ $x_1 = \frac{ - (- 1.35 \times 10^{- 3})+ \sqrt { 5.418 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.35 \times 10^{- 3})- \sqrt { 5.418 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 1.231 \times 10^{- 2} (Negative)$ $x_2 = 1.096 \times 10^{- 2}$ - The concentration can't be negative, so $[H_3O^+]$ = $x_2$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.01096)$ $pH = 1.96$