Question

3 attempts left Check my work Be sure to answer all parts. A volume of 66 mL of 0.080 M NaF is mixed with 22 mL of 0.20 M Sr(NO3)2. Calculate the concentrations of the following ions in the final solution.(Ksp for SrF2 2.0 x 10-10) [NO3-] [sx*] [F1

Answer 1

SOLN 1

NaF molarity = millimoles / total volume

millimoles = molarity * volume in mL

= 66 x 0.080 / (66 + 22)

= 0.060 M

Sr(NO3)2 molarity = 22 x 0.20 / (66 + 22)

= 0.050 M

Sr(NO3)2 -----> Sr2+ + 2NO3^-

[NO3-] = 2 x 0.050 = 0.100 M

[Na+] = 0.060 M

[NO3-] = 0.10 M

Sr+2 + 2F- ----------------> SrF2   

1    2

0.05 0.06

limiting reagent is F-

Sr+2 left = 0.05 - 0.06 / 2 = 0.02 M

[Sr⁺²] = 0.020 M

Ksp = [Sr+2] [F-]^2

2.0 x 10^-10 = 0.02 [F-]^2

[F-] = 1.0 x 10^-4 M

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