A volume of 75mL of 0.060M NaF is mixed with 25mL of 0.15M Sr(NO3)2. What compou
ID: 934822 • Letter: A
Question
A volume of 75mL of 0.060M NaF is mixed with 25mL of 0.15M Sr(NO3)2. What compound, if any, will precipitate? Calculate the concentrations in the final solution of NO3-, Na+, Sr2+, & F-.I have the answers just not sure how to get there. Can you please show all work. Thanks! A volume of 75mL of 0.060M NaF is mixed with 25mL of 0.15M Sr(NO3)2. What compound, if any, will precipitate? Calculate the concentrations in the final solution of NO3-, Na+, Sr2+, & F-.
I have the answers just not sure how to get there. Can you please show all work. Thanks!
I have the answers just not sure how to get there. Can you please show all work. Thanks!
Explanation / Answer
moles Na+ = moles F- = 0.075 L x 0.060 M =0.0045
moles Sr2+ = 0.025 x 0.15 M= 0.0038
moles NO3- = 2 x 0.038=0.076
total volume = 100 mL = 0.100 L
[Na+]=0.0045 / 0.100 L = 0.045 M
[NO3-]= 0.076 / 0.100 L= 0.76 M
[F-] = 0.0045 / 0.100 L = 0.045 M
[Sr2+]=0.0038 / 0.100 L = 0.038 M
the precipitation reaction
Sr2+ + 2 F- ---> SrF2
requires two moles F- per one mole Sr2+ so Sr2+ is present in excess.
If we assume it drives all F- out of solution then to remove 0.038/2 = 0.019 M Sr2+
[Sr2+]= 0.038 - 0.019 =0.019 M
at equilibrium
[F-] = 2x
[Sr2+]= 0.019+x
2.0 x 10^-10 = ( 0.019+x)( 2x)^2
neglecting x where it adds to 0.019
2.0 x 10^-10 = 0.019 ( 2x)^2
x = 5.1 x 10^-5 M
2x = 1.0 x 10^-4 M = [F-]
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