Question

Consider a 1.0 L solution which is 0.40 M CH₃CO₂H and 0.2 M CH₃CO₂Na (K_a for CH₃CO₂H = 1.8 x 10⁻⁵). Calculate the pH of the original soultion, the pH after 0.10 mol of HCl is added to the original solution, and the pH after 0.20 mol of NaOH is added to the original solution. Calculate all the pH values to two decimal places. Assume no volume change on addition. Please explain and /or show work please.

Answer 1

Answer - We are given, volume of solution = 1.0 L

Molarity of CH₃CO₂H ₂ = 0.40 M , molarity of CH₃CO₂Na = 0.20 M

We know, Ka for CH₃CO₂H = 1.8*10^-5

pH after adding 0.10mol of HCl –

First we need to calculate moles of acid and it conjugate base

Moles of CH₃CO₂H ₂ = 0.40 * 1.0 L = 0.40 moles

Moles of CH₃CO₂Na =0.20 * 1.0 L = 0.20 moles

When new added acid, HCl then it increases then moles of acid and decrease the moles of base

So new moles of acid and its conjugate base is

Moles of CH₃CO₂H ₂ = 0.40 + 0.10 = 0.50 moles

Moles of NaC₂H3O₂ = 0.20 – 0.10 = 0.10 moles

New molarity,

[CH₃CO₂H ₂ ] = 0.50 moles / 1.0 L = 0.50 M

[CH₃CO₂Na] = 0.10 moles / 1.0 L = 0.10 M

Now using the Henderson Hasselbalch equation-

pH = pKa + log [conjugate base] / [acid]

so first we need to calculate pKa from the Ka

pKa = - log Ka

        = - log 1.75 *10^-5

         = 4.76

pH = 4.76 + log 0.10 / 0.50

     = 4.76 + (-0.698)

= 4.06

When new added acid, NaOH then it increases then moles of base and decrease the moles of acid

So new moles of acid and its conjugate base is

Moles of CH₃CO₂H ₂ = 0.40 -0.20 = 0.20 moles

Moles of NaC₂H3O₂ = 0.20 +0.20 = 0.40 moles

New molarity,

[CH₃CO₂H ₂ ] = 0.20 moles / 1.0 L = 0.20 M

[CH₃CO₂Na] = 0.40 moles / 1.0 L = 0.40 M

Now using the Henderson Hasselbalch equation-

pH = pKa + log [conjugate base] / [acid]

pH = 4.76 + log 0.40 / 0.20

     = 4.76 + 0.3010

= 5.06