Why does the pH before the equivalence point of a titration depend on the initial concentration only when the acid is strong?
To understand what is happening with the two sets of titration curves, start by considering the titration of acetic acid ("HOAc" for short) with sodium hydroxide solution. The titration curves are shown below:



In the figure, the mutual concentrations are simply the concentrations of the reactants. Thus, 0.1 M HOAc is titrated with 0.1 M NaOH, 0.01 M HOAc is titrated with 0.01 M NaOH, and so on.

Acetic acid is a monoprotic weak acid, with $\mathrm{p}K_\mathrm{a}=4.756$. The titration curves for the three highest mutual concentrations almost coincide at the half-equivalence points, i.e, at the center of the buffer region. This is as expected from the Henderson-Hasselbalch equation, which ignores the autoionization of water. However, the three lower concentrations demonstrate that the titration curves do not coincide at the half-equivalence points: as the concentrations decrease, the pH values at the half-equivalence pints move toward the assumed pH of 7 for pure water.

As the concentrations are lowered, the effect of water's auto-ionization can no longer be neglected and the Henderson-Hasselbalch approximation is no longer viable. The calculated pH values at the half-equivalence points are shown in the second column of the table below:



Note that for 0.1 M, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $, as expected. Even at 0.001 M, the approximation is not bad. However, as concentrations continue to decline, the pH at the half-equivalence points simply move even closer to that assumed for pure water.

Now consider the titration of hydrochloric acid with sodium hydroxide solution. The titration curves are shown below:



As shown in the figure and in the third column of the table, the pH values at the half-equivalence points are one pH unit apart until the concentration is down around $\mathrm{1 \mu M}$.  Then, as concentrations continue to decline, the pH at the half-equivalence points must invariably move toward that of pure water.

For hydrogen ion concentrations high enough to neglect the auto-ionization of water, the pH at the half-equivalence point is easy to calculate. First, divide the initial hydrogen ion concentration by 2, to account for the fact that half of the number of hydrogen ions, from the 50 mL of starting HCl solution, were neutralized by addition of 25 mL of equimolar NaOH solution. Then multiply the hydrogen ion concentration by 2/3, because the solution volume increased from 50 mL to 75 mL. The net result is that the hydrogen ion concentration, at the half-equivalence point, is 1/3 of the initial value. Then take the negative logarithm base 10, as usual:

$$\mathrm{pH} = -\log{({\ce{[H+] \times (1/2) \times (2/3)) = - \log{(\ce{[H+]/3)}}}} }$$

If $\ce{[H^+]}$ = 0.1 M, then pH = 1.477. N.B. Significant figures were ignored in this answer, so 0.1 M is treated as exact, etc.