# Chemistry

A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml. 25.00 ml of the solution is taken out and is titrated to a neutral endpoint with 0.10 M NaOH. The titrated portion is then mixed with the remaining untitrated portion and the pH of the mixture is measured.

Mass of acid weighed out (grams) 0.732
Volume of NaOH required to reach endpoint: (ml) 17.8
pH of the mixture (half neutralized solution) 3.54

1) What is the molarity of the original acid solution?

2) What is the molecular weight of the acid?

Work:
1) (0.10 mmol / mL ) x (17.8 mL) = 1.78 mmol NaOH
1.78mmol*25 ml=44.5mmol Acid
44.5 M/50=.89-- This answer is wrong.

2).732 g /.0445 mol = 16.45 g/mol--Wrong

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1. Many thanks for showing your work. It helps me spot the problem right off. BTW, I assume this is a monoprotic acid. Your error is in step 2. 1.78 mmols NaOH is correct. That means mmols HA is 1.78.
You took 25 mL for that titration and M = mmols/mL; therefore, 1.78 mmols/25 mL = ?M
For #2, mols = grams/molar mass or
molar mass = grams/mols. You had 0.732/2 grams titrated in the 25.00 mL and you had 0.0178 mols in that 25 mL; therefore, 0.365/0.0178 = about 20 molar mass.
One point I should bring up about the problem. Not only did the problem not state how many H ions were there, it also said the 25 mL was titrated to the neutral point. To me that means pH = 7.0 and if that's true I don't think all of the weak acid was titrated. But that's not part of the problem. I think it's just stated poorly.

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