## Chemistry: Atoms First (2nd Edition)

$pH = 2.681$
1. Calculate the molar mass: 1.01* 1 + 12.01* 9 + 1.01* 7 + 16* 4 = 180.17g/mol 2. Calculate the number of moles - Note, each tablet has 0.325g, and we are dissolving 2 tablets, so the total mass is 0.325 * 2 = 0.65g $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.65}{ 180.17}$ $n(moles) = 3.608\times 10^{- 3}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 3.608\times 10^{- 3}}{ 0.237}$ $C(mol/L) = 0.01522$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_9H_7{O_4}^-] = x$ -$[HC_9H_7O_4] = [HC_9H_7O_4]_{initial} - x = 0.01522 - x$ For approximation, we consider: $[HC_9H_7O_4] = 0.01522M$ 5. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_9H_7{O_4}^-]}{ [HC_9H_7O_4]}$ $Ka = 3.3 \times 10^{- 4}= \frac{x * x}{ 0.01522}$ $Ka = 3.3 \times 10^{- 4}= \frac{x^2}{ 0.01522}$ $5.023 \times 10^{- 6} = x^2$ $x = 2.241 \times 10^{- 3}$ Percent dissociation: $\frac{ 2.241 \times 10^{- 3}}{ 0.01522} \times 100\% = 14.72\%$ %dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 3.3 \times 10^{- 4}= \frac{x^2}{ 0.01522- x}$ $5.023 \times 10^{- 6} - 3.3 \times 10^{- 4}x = x^2$ $5.023 \times 10^{- 6} - 3.3 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 3.3 \times 10^{- 4})^2 - 4 * (-1) *( 5.023 \times 10^{- 6})$ $\Delta = 1.089 \times 10^{- 7} + 2.009 \times 10^{- 5} = 2.02 \times 10^{- 5}$ $x_1 = \frac{ - (- 3.3 \times 10^{- 4})+ \sqrt { 2.02 \times 10^{- 5}}}{2*(-1)}$ or $x_2 = \frac{ - (- 3.3 \times 10^{- 4})- \sqrt { 2.02 \times 10^{- 5}}}{2*(-1)}$ $x_1 = - 2.412 \times 10^{- 3} (Negative)$ $x_2 = 2.082 \times 10^{- 3}$ - The concentration can't be negative, so $[H_3O^+]$ = $x_2$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 2.082 \times 10^{- 3})$ $pH = 2.681$