Question

11.

A 150.0 mL solution of 2.713 M strontium nitrate is mixed with 220.0 mL of a 2.401 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate.

mass: g

Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration.

[Na+] M

[NO−3] M

[Sr2+] M

[F−] M

Answer 1

You get the answer by following these steps.

moles of Sr(2+) = 0.15 L * 2.713 mol/L = 0.40695 moles moles of F(-) = 0.22 L * 2.401 mol/L = 0.52822 moles. It takes 2 mol F- to precipitate 1 mol of Sr(2+).Limiting reagent is F- ,

Mass of SrF2 = ( 0.52822moles) / 2 * ( 125.62g/mol) = 33.17 g

Na+ and NO3 - are spectator ions. The final volume of resulting products is 150ml + 220ml = 370 ml (0.37 L). The concentrations of Na+ and NO3 - can be calculated as [Na+] : (220 ml) * ( 2.401M) = (370ml) * ( x M) x = ([Na+]) = 1.427 M similarly,

[NaNO3 -] = (406.95 moles) /370 ml = 1.099 M

Initial moles of Sr(2+) = 0.40695 moles moles Sr(2+) precipitate by F- = 0.52822 /2 moles = 0.264mol

1 mol Sr(2+) is precipitated by 2 moles of F-. Excess of Sr(2+) remains = 0.406 - 0.264 = 0.142 mols (unprecipitated).

Concentration of Sr(2+), [Sr(2+)] = 0.142/0.37 = 0.384 M To find concentration of [F-] we have to know the value of Ksp of SrF2. Ksp = ​​​​​​(Note: there are several values for Ksp. Use the value that given in your textbook)

  ​​​​​​ [Sr2+] = 0.384M

[F-] ^2 = ​​​​​​

[F-] =  

Concentration of [F-] may be 0 M.