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A solution was made by dissolving 0.580 g of an unknown monoprotic acid in water, and...

A solution was made by dissolving 0.580 g of an unknown monoprotic acid in water, and diluting the solution to a final volume of 25.00 mL. This solution was then titrated with 0.100 M NaOH. It took 36.80 mL of NaOH to reach the equivalence point, at which point the pH was 10.42.

a. Determine the molar mass of the unknown acid.

b. Calculate what the pH was after 18.40 mL of NaOH was added during the titration. Hint: the pH at the equivalence point will need to be used here to help you. Think about what is present and how to calculate the pH based on that.

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