a 0.400-g sample of toothpaste was boiled with a 50-mL solution containinga citr

Question

a 0.400-g sample of toothpaste was boiled with a 50-mL solution containinga citrate buffer and NaCl to extract the fluoride ion. After cooling, the solution is diluted to exactly 100 mL. The potential of an ion-selective electrode with Ag-AgCl(sat'd) reference electrode in 25.0-mL aliquot of the sample was found to be -0.1823V. Addition of 5.0 mL of a solution containing 0.00107 mg F-/mL caused the potential to change to -0.2446V. Calculate the mass percentage of F- in the sample.

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Explanation / Answer

E= E0+ 2.303 log[2F-]^2 = 2.87 + 2.303 X 2log 4 F-

-0.1823 = 2.87 + 2.303 log 4 [F-] =

-3.0523 / 2.303 X 2 = log 4[F-]

[F-] = 0.220 /4 = 0.055

so amount of F- = 0.0473 X 19 g/l = 1.05 g/L

so in 25mL = 0.026 g

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