A solid weak acid is weighed, dissolved in water and diluted to exactly 50.00 ml
ID: 1063975 • 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) 0.709 Volume of NaOH required to reach endpoint: (ml) 17.9 pH of the mixture (half neutralized solution) 3.29 Calculate the following 1) What is the pKa of the acid? 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) 0.709 Volume of NaOH required to reach endpoint: (ml) 17.9 pH of the mixture (half neutralized solution) 3.29 What is the molarity of the original acid solution? Answer: 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) 0.709 Volume of NaOH required to reach endpoint: (ml) 17.9 pH of the mixture (half neutralized solution) 3.29 What is the molecular weight of the acid? Answer:Explanation / Answer
Titration of acid solution with NaOH
AH + NaOH = NaA + H2O
Volume of acid solution V1= 25 mL, Molarity of acid solution = M1
Volume of NaOH V2 = 17.9 mL, Molarity of NaOH solution M2= 0.10 M
For neutralization;
M1V1 = M2V2
M1 = M2V2/V1 = 0.10 M x 17.9 mL/25 mL = 0.0716 M (Molarity of original solution)
Number of mili moles of NaA formed = number of mili moles of NaOH used = M2V2 = 0.10 M x 17.9 =1.79 m mole
Total volume after titration = 25 mL + 17.9 mL = 42.9 mL
Molarity of salt (NaA) solution = number of milimoles/volume (mL) = 1.79 /42.9 = 0.042 M
After mixing the two solution, the volume of half neutralized solution = 25 + 42.9 = 67.9 mL
Molarity of NaA before mixing M1 = 0.042 M, volume V1 = 42.9 mL
Molarity of NaA after mixing M2 = ? V2 = 67.9
M2 = M1V1/V2 = 0.042 M x 42.9 mL/mL67.9= 0.0265 M
Molarity of acid before mixing = 0.0716, volume = 25 mL
Molarity of acid after mixing = 0.0716 x 25/67.9 = 0.0264 M
Since this is a buffer solution, so, from the Henderson’s equation:
pH = pKa + log[salt]/[acid]
3.29 = pKa + log 1 (since [salt] = [acid] = 0.0265)
pKa = 3.29
Number of milimoles of acid present in original solution = molarity x volume (mL)
= 0.0716 M x 50 mL = 3.58 m mol
Number of moles = 3.58/1000 = 3.58 x 10-3
Number of moles (n) = weight/Mw
Mw = weight/n = 0.709 /(3.58 x 10-3) = 198.04
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