Answer
The pH of this solution is equal to 2.75.
Work Step by Step
1. Find the amount of moles in the 15.0 mL sample:
$ HC_2H_3O_2 $ : ( 1.008 $\times$ 4 )+ ( 12.01 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 60.05g/mol
$$15.0 \space mL \times \frac{1.05 \space g}{1 \space mL} \times \frac{1 \space mol}{60.05 \space g} = 0.262 \space mol$$
2. Find the concentration of acetic acid in the resulting solution:
$$\frac{0.262 \space mol}{1.50 \space L} =0.175 \space M$$
3. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.175 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.175 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
4. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H^+ ]}{[ HC_2H_3O_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_2H_3O_2 ]_{initial} - x}$$
5. Assuming $ 0.175 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.175 }$$
$x = 1.8 \times 10^{-3} $
6. Test if the assumption was correct:
$$\frac{ 1.8 \times 10^{-3} }{ 0.175 } \times 100\% = 1.0 \%$$
Percent ionization > 5%: Valid
7. The percent is less than 5%. Thus, it is correct to say that $x = 1.8 x 10^{-3} $
8. $$[H_3O^+] = x = 1.8 \times 10^{-3} $$
9. Calculate the pH:
$$pH = -log[H_3O^+] = -log( \sqrt{ 1.8 \times 10^{-5} \times 0.175 } ) = 2.75 $$
