Question

Strontium chloride and sodium fluoride react to form strontium fluoride and sodium chloride, according to the reaction shown.

SrCl2(aq)+2NaF(aq)⟶SrF2(s)+2NaCl(aq)
What volume of a 0.550 M NaF
solution is required to react completely with 675 mL
of a 0.680 M SrCl2 solution?

Volume:

Answer 1

To calculate the volume of a 0.550 M NaF solution required to react completely with 675 mL of a 0.680 M SrCl2 solution, we can use the formula M1V1 = M2V2 where M1 is the molarity of the NaF solution, V1 is the volume of the NaF solution, M2 is the molarity of the SrCl2 solution and V2 is the volume of the SrCl2 solution.

Rearranging this formula to solve for V1 gives:

V1 = (M2V2) / M1

Substituting the values given in the problem statement:

V1 = (0.680 M x 675 mL) / 0.550 M

V1 = 836 mL

Therefore, 836 mL of a 0.550 M NaF solution is required to react completely with 675 mL of a 0.680 M SrCl2 solution.

I hope this helps! Let me know if you have any other questions. :)