Question

A 170.0 mL solution of 2.623 M strontium nitrate is mixed with 200.0 mL of a 3.375 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration.

Answer 1

Answer:

The mass of the resulting strontium fluoride precipitate is 42.40 grams

[Na+]  =  0.675 moles / 0.370 L = 1.82 M

[NO3-] = (0.170L * 2.623) = (0.370L * C2)  C2 = 1.21 M

[Sr2+] = the unprecipated part = 0.1085 mol/0.370 L = 0.293 M

Explanation:

Step 1: Data given

Volume of 2.623 M strontium nitrate = 170.0 mL = 0.170 L

Volume of 3.375 M sodium fluoride = 200.0 mL = 0.200 L

Molar mass of SrF2 = 125.62 g/mol

Step 2: The balanced equation:

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

SrF2 will precipitate, NaNO3 will dissociate in 2Na+ + 2NO3-

Step 3: Calculate moles of Sr(NO3)2

moles Sr(NO3)2 = molarity * volume

moles Sr(NO3)2 = 2.623 M * 0.170 L

moles Sr(NO3)2 = 0.446 moles

Step 4: Calculate moles NaF

moles NaF = 3.375 M * 0.200 L

moles NaF = 0.675 moles

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 (0.675 moles).

Sr(NO3)2 is in excess. There will react 0.675/2 = 0.3375 moles (will precipitate)

There will remain 0.446 - 0.3375 = 0.1085 moles ( will not precipitate)

Step 6: Calculate moles of SrF2 produced

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 0.675 moles of NaF consumed, we produced 0.675/2 = 0.3375 moles of SrF2

Step 7: Calculate mass of SrF2 produced

Mass SrF2 = moles SrF2 * molar mass SrF2

Mass SrF2 = 0.3375 moles * 125.62 g/mol

Mass SrF2 = 42.40 grams

The mass of the resulting strontium fluoride precipitate is 42.40 grams

Step 8: Calculate moles Na+ and NO3-

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 0.675 moles of NaF, we have 0.675 moles of NaNO3

For 0.675 moles of NaNO3, we'll have 0.675 moles of Na+ and 0.675 moles of NO3-

Step 9: Calculate concentration of the ions

Total volume = 170+200 = 370 mL

[Na+]  =  0.675 moles / 0.370 L = 1.82 M

[NO3-] = (0.170L * 2.623) = (0.370L * C2)  C2 = 1.21 M

[Sr2+] = the unprecipated part = 0.1085 mol/0.370 L = 0.293 M

Since NaF is completely be consumed, and SrF2 will completely precipitate, it means there will no F- ion