Complete and balance each acid–base equation.
a. H2SO4(aq) + Ca(OH)2(aq) →
b. HClO4(aq) + KOH(aq) →
c. H2SO4(aq) + NaOH(aq) →
Solution: Here, we are going to calculate the molar mass of the metal hydroxide used in the above titration. Also, we are going to identify the alkali metal cation in the metal hydroxide used. Step1: a) Given, molarity of the hydrochloric acid used = 2.50 M Volume of the hydrochloric acid used = 17.0 mL = 17 / 1000 L = 0.017 L Therefore, number of moles of hydrochloric acid used = volume of the solution in litres X molarity = 0.017 L X 2.50 M = 0.0425 mol Step2: Since, alkali metal forms monohydroxide, 1 mole of HCl will be required to neutralize 1 mole of the alkali metal hydroxide. Thus, to neutralize 0.0425 mol of HCl, 0.0425 moles of alkali metal hydroxide will be required. Now, Number of moles of the metal hydroxide = mass of the metal hydroxide / molar mass Putting the values in the above equation, we get, 0.0425 mol = 4.36 g / molar mass Molar mass = 4.36 g / 0.0425 mol = 102.59 g/mol Thus, molar mass of the alkali metal hydroxide is 102.59 g/mol Step3: b) Molar mass of the hydroxyl group(OH) = (1 x atomic mass of O) + (1 x atomic mass of H) = (1 x 15.9994) + (1 x 1.0079) = 15.9994 + 1.0079 = 17.01 g/mol Therefore, molar mass of the alkali metal cation = molar mass of the alkali metal hydroxide - molar mass of the hydroxyl group = 102.59 g/mol - 17.01 g/mol = 85.58 g/mol Out of all the alkali metals, Rubidium has an atomic mass equal to 85.46 u. Thus, the alkali metal cation in the hydroxide is Rb . ------------------