First, let us solve for the concentration of HCl solution used in the titration.
Given:
- Volume of HCl = \( 20 \, \text{mL} \)
- Molarity of HCl = \( 0.5 \, \text{M} \)
- Volume of HCl used in titration = \( 35.5 - 25.5 = 10 \, \text{mL} \)
The number of moles of HCl reacted is:
\[
\text{Moles of HCl} = M \times V = 0.5 \, \text{M} \times \frac{10}{1000} \, \text{L} = 0.005 \, \text{mol}
\]
The balanced equation between calcium hydroxide and HCl is:
\[
\text{Ca(OH)}_2 + 2 \, \text{HCl} \rightarrow \text{CaCl}_2 + 2 \, \text{H}_2\text{O}
\]
Thus, 1 mole of \( \text{Ca(OH)}_2 \) reacts with 2 moles of HCl
The number of moles of calcium hydroxide required for the titration is:
\[
\text{Moles of Ca(OH)}_2 = \frac{0.005}{2} = 0.0025 \, \text{mol}
\]
Now, for the unknown solution of H2SO4, the balanced equation is:
\[
\text{Ca(OH)}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + 2 \, \text{H}_2\text{O}
\]
1 mole of \( \text{Ca(OH)}_2 \) reacts with 1 mole of H2SO4
The moles of calcium hydroxide required for 10 mL of unknown H2SO4 solution is 0.0025 mol (from the titration)
Finally, the concentration of the unknown H2SO4 solution is:
\[
\text{Molarity of H}_2\text{SO}_4 = \frac{0.0025 \, \text{mol}}{0.01 \, \text{L}} = 0.25 \, \text{M}
\]
The concentration is 0.25 M