Answer to Question #106320 in General Chemistry for Ariana

Question

0.472 M sulfuric acid was used to titrate a 0.388 M barium hydroxide solution. The initial volume reading on the acid buret was 34.42 mL and the final reading was 41.90 mL. If the initial reading on the base buret was 63.25 mL, what was the final volume reading on the base?

Answer 1

$H_2SO_4(aq) + Ba(OH)_2(aq) \rightarrow BaSO_4(s) + 2H_2O(l)$

1) Find the volume of acid solution:

$\Delta V = V_{final}-V_{initial} = 41.90-34.42 = 7.48 mL$

2) Find the moles of $H_2SO_4$ in this solution

$7.48\times10^{-3}(\frac{0.472 mol H_2SO_4}{1L})=3.53\times10^{-3}mol H_2SO_4$

3) Find moles of $Ba(OH)_2$

$3.53\times 10^{-3} moles H_2SO4 (\frac{1 mole Ba(OH)_2}{1 mole H_2SO_4})=3.53\times10^{-3} moles Ba(OH)_2$

4) Find the volume of $Ba(OH)_2$ used:

$3.53\times10^{-3} mol Ba(OH)_2(\frac{1L}{0.388 mol Ba(OH)_2})=9.098\times 10^{-3} L=9.098 mL$

5) Find the final volume reading on the base burret

$V_{final} = V_{initial} + \Delta V = 63.25 mL + 9.098 mL =72.35 mL$

Answer: 72.35 mL