
Bob F. answered 03/04/16
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I'll answer question 1 -- follow the same procedure and see if you can answer #2 and #3.
First, write the balanced equation:
2 HNO3 + Ba(OH)2 → Ba+2aq + NO3-aq + 2 H2O
Figure moles of nitric acid used: 15.05 mL HNO3 x (1 L / 1000 mL) x (1.082 mole/L) = 0.016 moles HNO3
Use the stoichiometric ratio in the eqn: 0.016 moles HNO3 x ( 1 mole Ba(OH)2/ 2 moles HNO3) =
8.142 x 10-3 moles Ba(OH)2
Figure the concentration of the Ba(OH)2: (8.142 x 10-3 moles Ba(OH)2 / 23.67 mL) x (1000 mL/L) = 0.344 M Ba(OH)2
check: 23.67 mL x (1 L / 1000 mL) x (0.344 mole/L) = 8.143 x 10-3 moles
So, the answer is that the concentration of the Ba(OH)2 soln is 0.344 M
Use the same approach for numbers 2 and 3.