Answer
$[H_3O^+] = 5.107 \times 10^{- 4}M$
$[C_6H_5{CO_2}^-] = 5.107 \times 10^{- 4}M$
$[C_6H_5{CO_2}H] = 4.074 \times 10^{-3}M$
$[OH^-] = 1.958 \times 10^{- 11}M$
$pH = 3.292$
Work Step by Step
1. Calculate the molar mass:
12.01* 6 + 1.01* 5 + 12.01* 1 + 16* 2 + 1.01* 1 ) = 122.13g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.56}{ 122.13}$
$n(moles) = 4.585\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 4.585\times 10^{- 3}}{ 1} $
$C(mol/L) = 4.585\times 10^{- 3}M$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_6H_5{CO_2}^-] = x$
-$[C_6H_5CO_2H] = [C_6H_5CO_2H]_{initial} - x = 4.585 \times 10^{- 3} - x$
For approximation, we consider: $[C_6H_5CO_2H] = 4.585 \times 10^{- 3}M$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5{CO_2}^-]}{ [C_6H_5CO_2H]}$
$Ka = 6.4 \times 10^{- 5}= \frac{x * x}{ 4.585\times 10^{- 3}}$
$Ka = 6.4 \times 10^{- 5}= \frac{x^2}{ 4.585\times 10^{- 3}}$
$ 2.935 \times 10^{- 7} = x^2$
$x = 5.417 \times 10^{- 4}$
Percent dissociation: $\frac{ 5.417 \times 10^{- 4}}{ 4.585\times 10^{- 3}} \times 100\% = 11.81\%$
%dissociation < 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 6.4 \times 10^{- 5}= \frac{x^2}{ 4.585 \times 10^{- 3}- x}$
$ 2.935 \times 10^{- 7} - 6.4 \times 10^{- 5}x = x^2$
$ 2.935 \times 10^{- 7} - 6.4 \times 10^{- 5}x - x^2 = 0$
$\Delta = (- 6.4 \times 10^{- 5})^2 - 4 * (-1) *( 2.935 \times 10^{- 7})$
$\Delta = 4.096 \times 10^{- 9} + 1.174 \times 10^{- 6} = 1.178 \times 10^{- 6}$
$x_1 = \frac{ - (- 6.4 \times 10^{- 5})+ \sqrt { 1.178 \times 10^{- 6}}}{2*(-1)}$
or
$x_2 = \frac{ - (- 6.4 \times 10^{- 5})- \sqrt { 1.178 \times 10^{- 6}}}{2*(-1)}$
$x_1 = - 5.747 \times 10^{- 4} (Negative)$
$x_2 = 5.107 \times 10^{- 4}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
Therefore:
$[H_3O^+] = 5.107 \times 10^{- 4}M$
$[C_6H_5{CO_2}^-] = 5.107 \times 10^{- 4}M$
$[C_6H_5{CO_2}H] = 4.585 \times 10^{-3} - 5.107 \times 10^{- 4} = 4.074 \times 10^{-3}M$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 5.107 \times 10^{- 4})$
$pH = 3.292$
7. Calculate the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 5.107 \times 10^{- 4} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 5.107 \times 10^{- 4}}$
$[OH^-] = 1.958 \times 10^{- 11}$
