Answer
$[H_3O^+] = 1.468 \times 10^{- 6}M$
$pH = 5.833$
Work Step by Step
1. Calculate the molar mass:
12.01* 6 + 1.01* 5 + 16* 1 + 1.01* 1 = 94.12g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 0.195}{ 94.12}$
$n(moles) = 2.072\times 10^{- 3}$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
$ C(mol/L) = \frac{ 2.072\times 10^{- 3}}{ 0.125} $
$C(mol/L) = 0.01657$
4. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [C_6H_5O^-] = x$
-$[C_6H_5OH] = [C_6H_5OH]_{initial} - x = 0.01657 - x$
For approximation, we consider: $[C_6H_5OH] = 0.01657M$
5. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_6H_5O^-]}{ [C_6H_5OH]}$
$Ka = 1.3 \times 10^{- 10}= \frac{x * x}{ 0.01657}$
$Ka = 1.3 \times 10^{- 10}= \frac{x^2}{ 0.01657}$
$ 2.155 \times 10^{- 12} = x^2$
$x = 1.468 \times 10^{- 6}$
Percent ionization: $\frac{ 1.468 \times 10^{- 6}}{ 0.01657} \times 100\% = 0.008856\%$
%ionization < 5% : Right approximation.
Therefore: $[H_3O^+] = [C_6H_5O^-] = x = 1.468 \times 10^{- 6}M $
And, since 'x' has a very small value (compared to the initial concentration): $[C_6H_5OH] \approx 0.01657M$
6. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.468 \times 10^{- 6})$
$pH = 5.833$
