A solution is made by mixing 15.0 g of sr (oh) 2 and 55.0 ml of 0.200 m hno3.

a. write a balanced equation for the reaction that occurs between the solutes.

b. calculate the concentration of each ion remaining in solution.

c. is the resultant solution acidic or basic?

Letter A: The balanced equation is:

Sr (OH) 2 (s) + 2 HNO3 (aq) → Sr (NO3) 2 (aq) + 2 H2O

Letter B: The solution is:

(15.0 g Sr (OH) 2) / (121.6358 g Sr (OH) 2/mol) = 0.12332 mol Sr (OH) 2

(0.0550 L) x (0.200 mol/L HNO3) = 0.011 mol HNO3

0.011 mole of HNO3 would respond entirely with 0.011 x (1/2) = 0.0055 mole of Sr (OH) 2 but there is more Sr (OH) 2 existing than that, so Sr (OH) 2 is in excess and HNO3 is the limiting reactant.

(0.011 mol HNO3) x (1/2) = 0.0055 mol Sr (NO3) 2

Since the NO3{-} ions continued in solution during, the concentration of the NO3{-} ions didn't change during the reaction, so they are still 0.200 M.

The concentration of Sr{+2} ions is (0.0055 mol Sr (NO3) 2) / (55.0 mL) = 0.100 M

Since HNO3 is the limiting reactant, no H{+} ions continue at the end of the reaction.

Letter C: Since all of the acid reacted and there was still Sr (OH) 2 left over, the resulting solution would be basic.