Question

A 160.0 mL solution of 2.813 M strontium nitrate is mixed with 215.0 mL of a 2.900 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

Mass = 39.2 g

Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a zero for the concentration. [Na+] = ?

[Sr2+]=0.37 M

[F-]= 0 M

[NO3-]= ?

Answer 1

The balanced equation is

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

160x2.813= 450.08 215x2.90=623.5 0 0 initial mmoles

450.08 - (623.5/2) 0 623.5/2 623.5 mmoles after reaction

138.33 0 =311.75 623.5 mmoles after reaction

Thus the mas of Strontim fluoride precipitated = moles x molar mass

= 311.75 x10^-3 x 125.62 g/mol

= 39.162 g

Assming complete precipitation , after precipitation

[Sr+2] = mmoles/ total volume = 138.33 / (160+215)

= 138.33/375

= 0.3688 M

[Na+] = mmoles/ total volume = 623.5/375

= 1.6627 M

All the Na+ ions are form sodium fluoride solution, and are spectator ions thus the mmoles of Na+ = mmoles of NaF2 = 623.5

[F-] = 0M

[NO3-] = mmoles/total volume = 450.08/375

= 1.2 M

The nitrate ion is the specator ion which does not participate/precipitate in the reaction.

All the nitrate comes from strontium nitrate so the mmoles of nitrate in solution (total) = 450.08 and the volume of solution = 160+ 215 = 375 mL