Question

A 180.0 mL solution of 2.172 M strontium nitrate is mixed with 220.0 mL of a 2.634 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. and then Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration.

Answer 1

Sr(NO3)2(aq) + 2NaF(aq) ----> SrF2(s) + 2NaNO3(aq)

1 mol Sr(NO3)2 = 2 mol NaF = 1 mol SrF2 = 2 mol NaNO3

No of mol of Sr(NO3)2 = 180*2.172/1000 = 0.391 mol

No of mol of NaF = 220*2.634/1000 = 0.58 mol

limiting reactant = NaF

No of mol of SrF2 formed = 0.58*1/2 = 0.29 mol

amount of SrF2 formed = 0.29*125.62 = 36.43 g

Concentration of Na^+ = n/V = 0.58*1000/(180+220) = 1.45 M

Concentration of NO3^- = 0.391*2*1000/(180+220) = 1.955 M

Concentration of excess S^2+ = (0.391 - (0.58/2))*1000/(180+220) = 0.2525 M