Question

A solution of F– is prepared by dissolving 0.0980 ± 0.0005 g of NaF (molecular weight = 41.989 ± 0.001 g/mol) in 157.00 ± 0.09 mL of water. Calculate the concentration of F– in solution and its absolute uncertainty.

Answer 1

moles of F - = mass / molar mass = 0.0980 / 41.989 = 0.002334

[F-] = moles / volume = 0.002334 x 1000 / 157 = 0.01487 M

mass of NaF = 0.0005 x 100 / 0.0980 = 0.51 %

MW of NaF = 0.001 / 41.989 ) x 100 = 0.0024 %

volume = 0.09 x 100 / 157 = 0.057 %

% relative uncertainity = sqrt { (0.51)^2 + (0.0024)^2 + (0.057)^2 }

= 0.51 %

absolute uncertainity = 0.51 x 0.01487 / 100

= 7.6 x 10^-5 M

concentration of F - = 0.01487 M

absolute uncertainty =  7.6 x 10^-5 M