Answer
$pH = 2.75$
Work Step by Step
1. Calculate the molar mass:
1.01* 1 + 12.01* 2 + 1.01* 3 + 16* 2 = 60.06g/mol
** We have to convert the concentration from g/ml to mol/L, so:
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{1.05}{ 60.06}$
$n(moles) = 0.0175$
3. Find the concentration in mol/L:
$C(mol/L) = \frac{n(moles)}{volume(L)}$
** 1ml = 0.001L
$ C(mol/L) = \frac{ 0.0175}{ 0.001} $
$C(mol/L) = 17.5$
4. Find the concentration after the dilution:
$C_1 * V_1 = C_2 * V_2$
$17.5*0.015= C_2 *1.5$
$0.262 = C_2 *1.5$
$C_2 =0.175$
5. We have those concentrations at equilibrium:
-$[H_3O^+] = [C_2H_3{O_2}^-] = x$
-$[HC_2H_3O_2] = [HC_2H_3O_2]_{initial} - x = 0.175 - x$
For approximation, we consider: $[HC_2H_3O_2] = 0.175M$
6. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][C_2H_3{O_2}^-]}{ [HC_2H_3O_2]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.175}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.175}$
$ 3.15 \times 10^{- 6} = x^2$
$x = 1.77 \times 10^{- 3}$
Percent dissociation: $\frac{ 1.77 \times 10^{- 3}}{ 0.175} \times 100\% = 1.01\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [C_2H_3{O_2}^-] = x = 1.77 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HC_2H_3O_2] \approx 0.175M$
7. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.77 \times 10^{- 3})$
$pH = 2.75$
