Question

Question 5 (1 point)
0.378 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then
25.00 mL of this solution is transferred to an Erlenmeyer flask and is titrated with
0.1041 M NaOH. The initial burette reading is 3.19 mL; when the titration endpoint
is reached, the final burette reading is 39.18 mL. How many moles of diprotic acid
are neutralized during the titration? Provide your answer in decimal form (e.g. 0.123)
to the correct number of significant figures.

Answer 1

Answer:

0.00125 moles H₃X

Solution and Explanation:

In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.

Volume of NaOH used = final burette reading - initial burette reading

                                      = 39.18 ml - 3.19 ml

                                      = 35.99 ml or 0.03599 L

Step 1: Moles of NaOH used

Number of moles = Molarity × Volume

Molarity of NaOH = 0.1041 M

Moles of NaOH = 0.1041 M × 0.03599 L

                          = 0.00375 mole

Step 2: Balanced equation for the reaction between triprotic acid and NaOH

The balanced equation is;

H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)

Step 3: Moles of the triprotic acid (H₃X used

From the balanced equation;

1 mole of the triprotic acid reacts with 3 moles of NaOH

Therefore; the mole ratio of H₃X to NaOH is 1 : 3.

Therefore;

Moles of Triprotic acid = 0.00375 mole ÷ 3

                                     = 0.00125 moles

Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.