A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 30.0 mL of LiOH. A)0.86 B)1.21 C)2.00 D)1.12 E)2.86

Question

Quizplus · Accepted Answer

First, let's write out the balanced chemical equation for the reaction:

HClO + LiOH → LiClO + H2O

Since the acid and base are not equally strong, we will need to use the equation for calculating the pH of a weak acid-strong base titration:

pH = pKa + log([base]/[acid])

We can find the pKa for HClO by using the Ka value:

Ka = [H+][ClO-]/[HClO]
1.3 × 10^-8 = x^2 / (0.18 - x)
where x is the concentration of [H+]
x = 3.14 × 10^-4

pKa = -log(Ka) = 7.89

Before we start the titration, the [HClO] is 0.018 M and the [LiOH] is 0. If we add 30.0 mL of 0.27 M LiOH, we will have a total volume of 130.0 mL and a total amount of LiOH added of (30.0 mL)(0.27 M) = 0.0081 moles.

The balanced chemical equation tells us that the acid and base react in a 1:1 ratio. Therefore, the moles of HClO originally present will now equal the moles of LiOH added:

moles of HClO = 0.018 mol/L × 0.100 L = 0.0018 moles
moles of LiOH added = 0.0081 moles
moles of HClO left = 0.0081 moles

We can set up an ICE table to calculate the equilibrium concentrations of the acid and base:

HClO + LiOH → LiClO + H2O
I    0.0081 mol    0 mol        0 mol       0 mol
C   -0.0081 mol   -0.0081 mol  +0.0081 mol +0.0081 mol
E   0             0            0.0081 mol  0.0081 mol

The equilibrium concentration of HClO is now 0 M, and the equilibrium concentration of LiClO is 0.0081 mol/0.13 L = 0.0628 M. We can use the pH equation to solve for the pH:

pH = pKa + log([base]/[acid])
pH = 7.89 + log(0.0628/0.018)
pH = 1.12

Therefore, the pH after the addition of LiOH is 1.12. The correct answer is D.

A 1000 ML Sample of 0