A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml
ID: 795958 • Letter: A
Question
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) 0.604
Volume of NaOH required to reach endpoint: (ml) 18.1
pH of the mixture (half neutralized solution) 3.49
Calculate the following
1) What is the molarity of the original acid solution
2) What is the molecular weight of the acid?
Explanation / Answer
(1) Let the acid be HA
HA + NaOH => NaA + H2O
Moles of HA = moles of NaOH = volume x molarity of NaOH
= 18.1/1000 x 0.10 = 0.00181 mol
Molarity = moles/volume of HA
= 0.00181/0.025
= 0.0724 M
(2) Moles of HA = volume x molarity of HA
= 50.00/1000 x 0.0724 = 0.00362 mol
Molecular weight = mass/moles of HA
= 0.604/0.00362
= 167 g/mol
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