## Chemistry (4th Edition)

$pH \approx 2.00$
1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [A^-] = x$ -$[HA] = [HA]_{initial} - x = 0.1M - x$ For approximation, we consider: $[HA] = [HA]_{initial} = 0.1M$ ** HA = Acid, A- = Conjugate base. 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][A^-]}{ [HA]}$ $Ka = 7.1 \times 10^{- 4}= \frac{x * x}{ 0.15}$ $Ka = 7.1 \times 10^{- 4}= \frac{x^2}{ 0.15}$ $1.06 \times 10^{- 4} = x^2$ $x = 1.03 \times 10^{- 2}$ 5% test: $\frac{ 1.03 \times 10^{- 2}}{ 1.5 \times 10^{- 1}} \times 100\% = 6.87\%$ %ionization < 5% : Inappropriate approximation, so we have to consider: $[HA] = [HA]_{initial} - x$ $Ka = 7.1 \times 10^{- 4}= \frac{x^2}{ 0.15- x}$ $1.06 \times 10^{- 4} - 7.1 \times 10^{- 4}x = x^2$ $1.06 \times 10^{- 4} - 7.1 \times 10^{- 4}x - x^2 = 0$ Bhaskara: $\Delta = (- 7.1 \times 10^{- 4})^2 - 4 * (-1) *( 1.06 \times 10^{- 4})$ $\Delta = 5.04 \times 10^{- 7} + 4.26 \times 10^{- 4} = 4.26 \times 10^{- 4}$ $x_1 = \frac{ - (- 7.1 \times 10^{- 4})+ \sqrt { 4.26 \times 10^{- 4}}}{2*(-1)}$ or $x_2 = \frac{ - (- 7.1 \times 10^{- 4})- \sqrt { 4.26 \times 10^{- 4}}}{2*(-1)}$ $x_1 = - 1.06 \times 10^{- 2} (Negative)$ $x_2 = 9.97 \times 10^{- 3}$ $[H_3O^+] = x = 9.97 \times 10^{- 3}$ 3. Find the pH: $pH = -log[H_3O^+]$ $pH = -log( 9.97 \times 10^{- 3})$ $pH = 2.00$