Question

Which has maximum molarity of H3O+ when there is 100% ionisation?
(a) 9.8 g H2SO4 in 100 ml solution
(b) 9.8 g H3PO4 in 100 ml solution
(c) 9.8 g H2SO3 in 100 ml solution
(d) 9.8 g H3PO2 in 100 ml solution
(Pls answer with steps)

Answer 1

Explanation:

$[Molarity]=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{volume of solution in L}}$

(a) 9.8 g $H_2SO_4$ in 100 ml solution.

Volume of the solution = 100 mL = 0.1 L

$[H_2SO_4]=\frac{9.8 g}{98 g/mol\times 0.1 L}= 1 mol/L$

1 mole of $H_2SO_4$ gives 2 moles of $H_3O^+$ on ionization.

$[H_3O^+]=2\times [H_2SO_4]=2\times 1 mol/L = 2.000 mol/L$

(b) 9.8 g $H_3PO_4$ in 100 ml solution.

Volume of the solution = 100 mL = 0.1 L

$[H_3PO_4]=\frac{9.8 g}{98 g/mol\times 0.1 L}= 1.000 mol/L$

1 mole of $H_3PO_4$ gives 3 moles of $H_3O^+$ on ionization.

$[H_3O^+]=3\times [H_3PO_4]=3\times 1.000 mol/L = 3.000 mol/L$

(c) 9.8 g $H_2SO_3$ in 100 ml solution.

Volume of the solution = 100 mL = 0.1 L

$[H_2SO_3]=\frac{9.8 g}{82 g/mol\times 0.1 L}= 1.1951 mol/L$

1 mole of $H_2SO_3$ gives 2 moles of $H_3O^+$ on ionization.

$[H_3O^+]=2\times [H_2SO_3]=2\times 1 mol/L = 2.3902 mol/L$

d) 9.8 g $H_3PO_2$ in 100 ml solution.

Volume of the solution = 100 mL = 0.1 L

$[H_3PO_2]=\frac{9.8 g}{82 g/mol\times 0.1 L}= 1.1951 mol/L$

1 mole of $H_2SO_3$ gives 3 moles of $H_3O^+$ on ionization.

$[H_3O^+]=3\times [H_2SO_4]=3\times 1.1951 mol/L = 3.5853 mol/L$