A solution of a theoretical triprotic acid was prepared by dissolving 5.826 g of
ID: 902600 • Letter: A
Question
A solution of a theoretical triprotic acid was prepared by dissolving 5.826 g of solid in enough DI water to make 500 Ml solution. 11.75 mL of a .530 M solution was required to titrate 20.00 mL of this acids solution1) What is the concentration of the acid solution?
2) What is the molar mass of the acid?
Last question is if 14.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution, what is the concentration of NaOH ?
A solution of a theoretical triprotic acid was prepared by dissolving 5.826 g of solid in enough DI water to make 500 Ml solution. 11.75 mL of a .530 M solution was required to titrate 20.00 mL of this acids solution
1) What is the concentration of the acid solution?
2) What is the molar mass of the acid?
Last question is if 14.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution, what is the concentration of NaOH ?
1) What is the concentration of the acid solution?
2) What is the molar mass of the acid?
Last question is if 14.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution, what is the concentration of NaOH ?
Explanation / Answer
The identity of the titrant is not given so i am assuming that NaOH is used to titrate against the acid solution.
The milliequivalents of the acid should be equal to the milliequivalents of the base
n1M1V1 = n2M2V2 for a titration of acid base & n1/n2 = basicity/acidity of the acid/base
3xM x 20 ml = 1x .530 x 11.75 mL
M of the acid solution =0.103 Moles/litre
Molarity of the acid solution = (Moles of the acid /volume of the solution)x1000
0.103 = {(5.826 g/Molar mass)/500} x 1000
Molar mass of the acid = 113.12 g/mole
--------------------------------------------------------------------------------------
last question
4.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution,
n1M1V1 = n2M2V2
4.18 x M X 1 = 2 x .6 X 15
Molarity of NaOH = 4.3 M
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.