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A volume of 75 mL of 0. 060 M NaF is mixed with 25 mL of 0. 15 M Sr(NO3)2. Calculate the concentrations in the final solution of NO3 - , Na , Sr2 , and F-. (Ksp for SrF2

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For a volume of 75 mL of 0. 060 M NaF is mixed with 25 mL of 0. 15 M Sr(NO3)2, the concentrations in the final solution is mathematically given as

  • [NO3-] = 0.075 M
  • [Na+] = 0.045 M
  • [Sr2+] = 0.015 M
  • [F-] = 0.00012 M

What are the concentrations in the final solution of NO3 - , Na , Sr2 , and F-.?

Generally, the equation for the molarity of NaF is mathematically given as

M NaF= millimoles / total volume

M NaF= = 75 x 0.060 / (75+ 25)

M NaF  = 0.045

Where, Sr(NO3)2 molarity = 25 x 0.15 / (75+ 25)

Sr(NO3)2 molarity = 0.0375 M

The Chemmical reaction of Sr and F is

Sr+2 + 2 F ---> SrF2  

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

2.0 x 10^-10 = 0.015 [F-]^2

[F-] = 1.2 x 10^-4 M

In conclusion,  the concentrations are

  • [NO3-] = 0.075 M
  • [Na+] = 0.045 M
  • [Sr2+] = 0.015 M
  • [F-] = 0.00012 M

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