Chemistry Forum

A titration analysis was done to determine the arsenic acid (H₃AsO₄) content of an impure sample. The titrant used is a 0.454 M NaOH solution.
a. Give the balanced equation of the reaction. (Hint: consider arsenic acid to fully dissociate; H₃AsO₄ → 3H+ + AsO₄₃-)

To prepare the analyte for titration, a 0.3000 g impure arsenic acid sample was dissolved in 25.00 mL water. This solution was then titrated with the NaOH standard.

b. Calculate the molarity of the 25.00 mL arsenic acid solution if the initial volume of the titrant in the burette is 44.32 mL, and the final volume at endpoint is 32.74 mL.

c. From the molarity of the arsenic acid, calculate the grams of pure arsenic acid present in the solution.

d. What is the percent purity of the arsenic acid sample?

My answers:
a. NaOH + H₃AsO₄ = Na₃AsO₄ + H₂O (Not sure)
b. 0.21 M (Not sure)
c. I don't know how
d. I know how but I have to get the answer in letter c first.

Please help me. Have a nice day!

My answers:

a)H₃AsO₄ + 3NaOH → Na₃AsO₄ + 3H₂O

b) 0.07 M H₃AsO₄ (25 mL)

c)0.249 g of Arsenic acid is present in the solution.

d) 82.9%

I cann't assure you that these answers are correct. Let the moderator decide it.

a. Is not balanced.
b. 11,58 ml NaOH was consumed, what is 5,257 mmol. This correspond to 1/3 = 1,752 mmol H₃AsO₄ what means 0,07 M
c = 248 mg
d = 82,9%

So, my answers to b), c) , d) are correct.
But you said equation in the answer a) is not balanced. How is that? Would you explain?

Thank you for the time and effort. I really appreciate it. But may I know how did you compute letter b and c? I really want to know how..

@Dhamnekar Winod, I think he's pertaining to my answer in letter a. Your answer is correct actually. I forgot to balance it, thank you for reminding me.

ChenBeier wrote: ↑Mon Mar 28, 2022 6:22 am a. Is not balanced.
b. 11,58 ml NaOH was consumed, what is 5,257 mmol. This correspond to 1/3 = 1,752 mmol H₃AsO₄ what means 0,07 M
c = 248 mg
d = 82,9%

I mean answer from Axel, not yours Dhamnekar winod

a. The balanced equation of the reaction is:
H₃AsO₄ + 3NaOH → Na₃AsO₄ + 3H₂O

One year later!!!!!!!!!!!!