Question

A student begins with 25 mL of a 0.434 M solution of HI and slowly adds a solution of 0.365 M NaOH.

1. What is the pH after 15.00 ml of NaOH solution has been added?

2. What is the pH after 40 ml of NaOH solution has been added?

Answer 1

You must know where the equivalence point is in order to know where you are on the titration curve.
mL acid x M acid = mL base x M base.
25*0.434 = mL x 0.365
mL base = estimated 30 mL so 15 mL is BEFORE the eq point and 40 mL is AFTER the eq point.
A. mols HI = M x L = ?
mols NaOH = M x L (0.015 x 0.365) = ?
Subtract, giving you mols HI, divide by total L (15 mL + 25 mL) to give M and convert to pH.

B. mols HI = M x L
mols NaOH= M x L
Subtract giving mols NaOH and divide by total L for M, then convert to pOH and pH.

Answer 2

1.mole of NaOH added = cv = 0.365x15e-3 = 5.48e-3 moles.

Reaction;

HI + NaOH --> NaI + H2O

pH = -log[H+]

the [H+] is the concentration of the acid as 15ml NaOH has been added.

the equation states that, 1 mole of NaOH neutralize 1 mole of HI. So, after adding 15ml, we introduce 5.48e-3 moles of NaOH in the solution. So, this mole should also neutralize 5.48e-3moles of HI.

concentration is moles/volume. The volume of acid is 25mL. with that, use -log[H+].

2. again, find the mole for 40ml of NaOH to find the mole of HI been neutralized. then use that mole to calculate the concentration..


hope that helpss..