The pH of saturated \(\operatorname{Sr}(\text { OH })_{2}(\text { aq })\) is found to be 13.12 A \(10.0 \mathrm{mL}\) sample of saturated \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is diluted to \(250.0 \mathrm{mL}\) in a volumetric flask. A \(10.0 \mathrm{mL}\) sample of the diluted \(\operatorname{Sr}(\mathrm{OH})_{2}(\mathrm{aq})\) is transferred to a beaker, and some water is added. The resulting solution requires \(25.1 \mathrm{mL}\) of a \(\mathrm{HCl}\) solution for its titration. What is the molarity of this HCl solution?

Since pH of saturated \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) is given as 13.12, we can calculate concentration of \(\operatorname{Sr}(\mathrm{OH})_{2}\) using formula for pOH and pH. The pOH is found by subtracting given pH from 14 (since pH + pOH = 14), which gives \(pOH = 14 - 13.12 = 0.88 \). The \(\operatorname{OH}^{-}\) concentration (molarity) is found by taking anti-log of \(-pOH\), so concentration of \(\operatorname{OH}^{-} = 10^{-0.88} \approx 0.132 \, mol \, L^{-1}\). As the formula of \(\operatorname{Sr}(\mathrm{OH})_{2}\) indicates, one molecule of it contains two \(\operatorname{OH}^{-}\) ions, the concentration of \(\operatorname{Sr}(\mathrm{OH})_{2}\) is half of the \(\operatorname{OH}^{-}\) molarity, which is \( 0.132/2 = 0.066 \, mol \, L^{-1}\).

The initial volume of the \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) solution is 10.0 mL and it is diluted to 250.0 mL. We can use the dilution formula: \(M1 \times V1 = M2 \times V2 \), where \(M1\) is initial molarity (0.066 mol/L), \(V1\) is initial volume (10.0 mL), \(V2\) is final volume (250.0 mL), and \(M2\) is the final molarity we need to find out. So, \(M2 = (M1 \times V1) / V2 = (0.066 \times 10.0) / 250.0 = 0.00264 \, mol \, L^{-1}\). This is the molarity of \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) in the final 250 mL solution.

In the titration, 10.0 mL of the diluted \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) solution was used, containing \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) of concentration \(0.00264 \, mol \, L^{-1}\). The stoichiometry of the reaction between \(\operatorname{Sr}\left(\mathrm{OH}\right)_{2}\) and \(\mathrm{HCl}\) is 1:2. Which is why, twice the moles of \(\mathrm{HCl}\) will be used for the titration. So, molarity of \(\mathrm{HCl}\) (let's denote it as \(M\)) is given by \((2 \times moles \, of \, \operatorname{Sr}\left(\mathrm{OH}\right)_{2}) / volume \, of \, HCl \, used\) which equals \((2 \times 0.00264 \, mol \, L^{-1} \times 0.01 \, L) / 0.0251 \, L = 0.021 \, mol \, L^{-1}\). This is the molarity of the \(\mathrm{HCl}\) solution.