A150.0 ml solution of 2.888 m strontium nitrate is mixed with 200.0 ml of a 3.076 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 [ sr 2 + ] = m [ f − ] = m

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

The mass of SrF2 precipitate formed in the reaction is 38.69 g

We have to first write down the balanced reaction equation as follows;

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

Next, we obtain the number of moles of each reactant;

Amount of Sr(NO3)2 = 150.0/1000 × 2.888  = 0.433 mols of Sr(NO3)2

Amount of NaF = 200.0/1000 × 3.076 = 0.615 moles of NaF

We have to obtain the limiting reactant. This is the reactant that yields the lowest number of moles of product.

For Sr(NO3)2:

1 mole of Sr(NO3)2 yields 1 mole of SrF2

0.433 mols of Sr(NO3)2 yields 0.433 mols of SrF2

For NaF;

2 moles of NaF yields  1 mole of SrF2

0.615 moles of NaF yields 0.615 × 1/2 = 0.308 moles of SrF2

Hence NaF is the limiting reactant.

Mass of  SrF2 formed =  0.308 moles of SrF2 × 125.62 g/mol = 38.69 g

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the answer is na (sodium)