A \(4.36-g\) sample of an unknown alkali metal hydroxide is dissolved in $100.0 \mathrm{~mL}$ of water. An acid-base indicator is added, and the resulting solution is titrated with \(2.50 \mathrm{MHCl}(a q)\) solution. The indicator changes color, signaling that the equivalence point has been reached, after \(17.0 \mathrm{~mL}\) of the hydrochloric acid solution has been added. (a) What is the molar mass of the metal hydroxide? (b) What is the identity of the alkali metal cation: $\mathrm{Li}^{+}, \mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+},\( or \)\mathrm{Cs}^{+} ?$

We are given that 17.0 mL of a 2.50 M HCl solution is added to reach the equivalence point. To find the moles of HCl added, we use this formula: moles of HCl = volume of HCl (in liters) × molarity of HCl (in mol/L). moles of HCl = (0.017 L) × (2.50 mol/L) = 0.0425 mol

At the equivalence point, the moles of HCl will be equal to the moles of the alkali metal hydroxide (MOH). Thus, moles of MOH = 0.0425 mol

Now that we have the moles of the alkali metal hydroxide, we can calculate its molar mass by dividing the given mass (4.36 g) by the moles of MOH (0.0425 mol). Molar mass of MOH = (4.36 g)/(0.0425 mol) = 102.588 g/mol

We are given the potential cations as Li+, Na+, K+, Rb+, and Cs+. We can find the molar mass of the hydroxides for each of these cations: - For LiOH: Molar mass = 6.939 (Li) + 15.999 (O) + 1.008 (H) = 23.946 g/mol - For NaOH: Molar mass = 22.990 (Na) + 15.999 (O) + 1.008 (H) = 39.997 g/mol - For KOH: Molar mass = 39.098 (K) + 15.999 (O) + 1.008 (H) = 56.105 g/mol - For RbOH: Molar mass = 85.468 (Rb) + 15.999 (O) + 1.008 (H) = 102.475 g/mol - For CsOH: Molar mass = 132.905 (Cs) + 15.999 (O) + 1.008 (H) = 149.912 g/mol The closest molar mass to our calculated value of 102.588 g/mol is RbOH with a molar mass of 102.475 g/mol. Thus, the identity of the alkali metal cation is Rb+. The final answers are: (a) Molar mass of the metal hydroxide is 102.588 g/mol, and (b) The identity of the alkali metal cation is \(\mathrm{Rb}^{+}\).