Question

what volume of a .55M NaF solution is required to reac completely with 301 mL of a .260 SrCl2 solution

Answer 1

To completely react with 301 mL of a 0.260 M SrCl2 solution, approximately 71.05 mL of a 0.55 M NaF solution is required.

Step-by-step explanation:

To determine the volume of the NaF solution required to react completely with the SrCl2 solution, we need to use the balanced chemical equation between NaF and SrCl2:
2NaF + SrCl2 → 2NaCl + SrF2

1. First, calculate the number of moles of SrCl2:

0.301 L x 0.260 M = 0.07826 mol SrCl2

1. Since the ratio of NaF to SrCl2 is 2:1, we need half the number of moles of NaF to completely react with SrCl2:

0.07826 mol SrCl2 x 0.5 = 0.03913 mol NaF

1. Next, calculate the volume of the NaF solution:

0.03913 mol NaF ÷ 0.55 M = 0.07105 L or 71.05 mL