Exam 1 (76).pdf - Question A 0.1276-g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with a 0.0633 M NaOH solution

# Exam 1 (76).pdf - Question A 0.1276-g sample of an unknown...

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Question: A 0.1276-g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with a 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid, (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid? Answer : a) Concentration of required to bring equivalence point = Volume of required to bring equivalence point = Number of moles of present = Hence number of moles of required to bring equivalence point is 0.00116. Let us suppose that HA be the monoprotic acid. According to the equation 1 mole of monoprotic acid is neutralized by 1 mole of base. Number of moles of monoprotic acid required to neutralize 0.00116mol Hence 0.00116 moles of monoprotic acid (HA) is neutralized by the 0.00116 moles    #### You've reached the end of your free preview.

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• Summer '16
• boer
• pH, 0.0633 M, 0.00116 moles, 0.1276 g, 0.1276-g
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