Question

An analytical chemist is titrating 66.8 mL of a 0.1400 M solution of trimethylamine (CH) N with a 0.3800 M solution of HIO3 ThepK,c trimethylamine is 4.19. Calculate the pH of the base solution after the chemist has added 26.5 mL of the HIO, solution to it Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HIO solution added Round your answer to 2 decimal places. PH

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Answer #1

base millimoles = 66.8 x 0.140 = 9.352

acid millimoles = 0.3800 x 26.5 = 10.07

acid millimoles are left = 10.07 -9.352 = 0.718

[H+] = 0.718 / (66.8 + 26.5) = 7.70 x 10^-3 M

pH = -log [H+]

pH = 2.11

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