## Chemistry: The Molecular Science (5th Edition)

$pH = 3.267$
1. Calculate the molar mass: 12.01* 3 + 1.01* 6 + 16* 3 = 90.09g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.056}{ 90.09}$ $n(moles) = 6.216\times 10^{- 4}$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 6.216\times 10^{- 4}}{ 0.25}$ $C(mol/L) = 2.486\times 10^{- 3}$ 4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_3H_5{O_3}^-] = x$ -$[C_3H_6O_3] = [C_3H_6O_3]_{initial} - x = 2.486 \times 10^{- 3} - x$ For approximation, we consider: $[C_3H_6O_3] = 2.486 \times 10^{- 3}M$ 5. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][C_3H_5{O_3}^-]}{ [C_3H_6O_3]}$ $Ka = 1.5 \times 10^{- 4}= \frac{x * x}{ 2.486\times 10^{- 3}}$ $Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 2.486\times 10^{- 3}}$ $3.73 \times 10^{- 7} = x^2$ $x = 6.107 \times 10^{- 4}$ Percent dissociation: $\frac{ 6.107 \times 10^{- 4}}{ 2.486\times 10^{- 3}} \times 100\% = 24.56\%$ %dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration: $Ka = 1.5 \times 10^{- 4}= \frac{x^2}{ 2.486 \times 10^{- 3}- x}$ $3.73 \times 10^{- 7} - 1.5 \times 10^{- 4}x = x^2$ $3.73 \times 10^{- 7} - 1.5 \times 10^{- 4}x - x^2 = 0$ $\Delta = (- 1.5 \times 10^{- 4})^2 - 4 * (-1) *( 3.73 \times 10^{- 7})$ $\Delta = 2.25 \times 10^{- 8} + 1.492 \times 10^{- 6} = 1.514 \times 10^{- 6}$ $x_1 = \frac{ - (- 1.5 \times 10^{- 4})+ \sqrt { 1.514 \times 10^{- 6}}}{2*(-1)}$ or $x_2 = \frac{ - (- 1.5 \times 10^{- 4})- \sqrt { 1.514 \times 10^{- 6}}}{2*(-1)}$ $x_1 = - 6.903 \times 10^{- 4} (Negative)$ $x_2 = 5.403 \times 10^{- 4}$ - The concentration can't be negative, so $[H_3O^+]$ = $x_2$ 6. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 5.403 \times 10^{- 4})$ $pH = 3.267$