Question

A solution is made by mixing 15.5 g of Sr(OH)2 and 40.0 mL of 0.230 M HNO3.

Part A: Write a balanced equation for the reaction that occurs between the solutes.

Part B: Calculate the concentration of OH− ion remaining in solution.

Part C: Calculate the concentration of Sr2+ ion remaining in solution.

Part D: Calculate the concentration of NO−3 ion remaining in solution.

Part E: Is the resultant solution acidic or basic?

Answer 1

Sr(OH)2 + 2HNO3 --------------> Sr(NO3)2 + 2 HOH

first calculate the moles of each reactant you have. For Sr(OH)2 use 15.5 g / 121.64 g/mol = 0.127 mol

For HNO3 it is C x V = 0.230 mol/L x 0.040 L = 0.0092

Now look at the mol ratio between Sr(OH)2 and HNO3. Can you see that it is 1 : 2? So for every mol of Sr(OH)2 you need TWICE as many mol of HNO3. Well do you have that many mol of HNO3?

Well you have 0.127 mol of Sr(OH)2, so twice as much would make it 0.127 x 2 or 0.254 mol

Well you only have 0.0092 mol of HNO3. Clearly not enough to use up all the Sr(OH)2. So HNO3 is the limiting reagent. The reaction will stop when 0.0092 mol of HNO3 is used up

Now how many moles of Sr(OH)2 will be used up? Well work in reverse. You only have 0.0092 mol of HNO3 but you only need half that amount for the Sr(OH)2. So 0.0092 / 2 = 0.0046 mol of Sr(OH)2 will be used.

Now all you have to do is find the concentration of that much Sr(OH)2. C = n / V so 0.0046 mol / 0.040 mL = 0.115 mol/L

That is the concentration of Sr(OH)2 remaining.

Since Sr(OH)2 --------------> Sr+2(aq) + 2OH-(aq), then the concentration of Sr+2 = 0.115 mol/L and the conc of OH- will be 2 x 0.115 mol/L or 0.230 mol/L

Half of the Sr(OH)2 compound remained so it is basic solution