Answer:
Mass SrF2 produced = 38.63 g SrF2 produced
[Na^+]: = 1.758 M
[NO3^-]: = 1.238 M
[Sr^2+] = 0.3589 M
[F^-] = 2.36*10^-5 M
Explanation:
Step 1: Data given
Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L
Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L
Step 2 : The balanced equation
Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2
Step 3: Calculate moles strontium nitrate
Moles Sr(NO3)2 = Molarity * volume
Moles Sr(NO3)2 = 2.888 M * 0.150 L
Moles Sr(NO3)2 = 0.4332 moles
Step 4: Calculate moles NaF
Moles NaF = 3.076 M * 0.200 L
Moles NaF = 0.6152 moles
It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.
Step 5: Calculate limiting reactant
For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3
NaF is the limiting reactant. It will completely be consumed (0.6152 moles).
Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles
Moles Sr^2+ precipitated by F^- = 0.3076
There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2
Moles Sr^2+ no precipitated (left over) = 0.1256 moles
Step 6: Calculate moles SrF2
For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3
For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2
Mass SrF2 produced: 0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced
Step 7: Calculate concentration of [Na+] and [NO3-]
Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:
[Na^+]: (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M
[NO3^-]: (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M
Step 8: Calculate [Sr^2+] and [F^-]
[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M
To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.
SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)
Ksp = [Sr^2+][F^-]²
2x10^-10 = (0.3589)(x)²
x² = 5.57*10^-10
x = [F^-] = 2.36*10^-5 M
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