Question

An 8.65 g sample of an unknown group 2A metal hydroxide is dissolved in 85.0 mL of water and titrated with 2.50 M HCl(aq). If it takes 56.9 mL of the acid to reach the end point of the titration
(a) What is the molar mass of the metal hydroxide?
(b) Which of the following is in the metal hydroxide: Ca2+, Sr2+, Ba2+?

Answer 1

Answer: (a)The molar mass of the metal is 121.66 g/mol.

              (b) The metal hydroxide is $Sr^{2+}$.

Explanation:

The equation for the reaction is as follows.

      $M(OH)_{2}+2HCl\rightarrow 2MCl_{2}+2H_{2}O$

(a) When 56.9 mL of the acid is required to reach end point of titration. Then,

Moles of acid = $Molarity \times Volume$

                        = $2.50 M \times \frac{56.9}{1000 L}$

                        = 0.14225 moles

Hence, the number of moles of hydroxide = 1 - 0.14225 = 0.0711 moles

It is given that mass of sample is 8.65 g, then calculate the molar mass of the sample as follows.

          $Molar mass = \frac{given mass}{no. of moles of hydroxide}$

                                     = $\frac{8.65 g}{0.0711 mole}$

                                     = 121.66 g/mol

The molar mass of the metal is 121.66 g/mol.

(b) The molar mass calculated is 121.66 g/mol. Therefore, calculate the molar mass of each given metal hydroxide as follows.

$Ca(OH)_{2}$ = (40.078 + 34) g/mol = 74.093 g/mol

$Sr(OH)_{2}$ = (87.66 + 34) g/mol = 121.66 g/mol

$Ba(OH)_{2}$ = (137.32 + 34) g/mol = 171.32 g/mol

Thus, it can be concluded that the metal in the metal hydroxide is $Sr^{2+}$.