75 ml of .060M NaF is added to 25ml of .15M SR(NO3)2. What is the F- concentrati

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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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