Answer
The titration curve with the lower initial pH is the one for the $0.100 \space M \space HCl$ solution.
Work Step by Step
Since $HCl$ is a strong acid, the concentration of hydronium ion $H_3O^+$ is greater than the same in a $HF$ (weak acid) solution with the same acid concentration.
With a greater $[H_3O^+]$, it will have a lower pH value.
$0.100 \space M \space HCl$:
Initial pH = -log(0.100) = 1.00
$0.100 \space M \space HF$:
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
$HF(aq) + H_2O(l) \lt -- \gt F^-(aq) + H_3O^+(aq) $
\begin{matrix} & [HF] & [H_3O^+] & [F^-] \\ Initial & 0.1 & 0 & 0 \\ Change & -x & +x & +x\\ Equil & 0.1 -x & x & x \end{matrix}
For approximation, we are going to consider $0.1 - x \approx 0.1$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][F^-]}{ [HF]}$
$Ka = 3.5 \times 10^{- 4}= \frac{x * x}{ 0.1}$
$Ka = 3.5 \times 10^{- 4}= \frac{x^2}{ 0.1}$
$x^2 = 0.1 \times 3.5 \times 10^{-4} $
$x = \sqrt { 0.1 \times 3.5 \times 10^{-4}} = 5.9 \times 10^{-3} $
Percent dissociation: $\frac{ 5.9 \times 10^{- 3}}{ 0.1} \times 100\% = 5.9\%$
%dissociation > 5% : Inappropriate approximation, so, we will have to consider the '-x' in the acid concentration:
$Ka = 3.5 \times 10^{-4} = \frac{x^2}{0.1-x}$
$(3.5 \times 10^{-4} \times 0.1) - 3.5 \times 10^{-4}x = x^2$
$(3.5 \times 10^{-4} \times 0.1) - 3.5 \times 10^{-4}x - x^2 = 0$
Bhaskara:
$x_1 = \frac{ - (- 3.5 \times 10^{-4})+ \sqrt {(3.5 \times 10^{-4})^2 - 4(-1)(3.5 \times 10^{-4} \times 0.1)}}{2*(-1)}$
or
$x_2 = \frac{ - (- 3.5 \times 10^{-4})- \sqrt {(3.5 \times 10^{-4})^2 - 4(-1)(3.5 \times 10^{-4} \times 0.1)}}{2*(-1)}$
$x_1 = - 6.1 \times 10^{- 3} (Negative)$
$x_2 = 5.7 \times 10^{- 3}$
- The concentration can't be negative, so $[H_3O^+]$ = $x_2$
3. Calculate the pH value:
$pH = -log[H_3O^+]$
$pH = -log( 5.7 \times 10^{- 3})$
$pH = 2.24$
