Question

For the next three problems, consider 1.0 L of a solution which is 0.6 M HC₂H₃O₂ and 0.2 M NaC₂H₃O₂ (K_a for HC₂H₃O₂ = 1.8 x 10^-5). Assume 2 significant figures in all of the given concentrations so that you should calculate all of the following pH values to two decimal places.

1. Calculate the pH of this solution.

2. Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl.

3. Calculate the pH after 0.20 mol of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH.

Answer 1

Solution :-

Given solution is the solution of weak acid and its conjugate base therefore its buffer solution

Using the Henderson equation we can calculate its pH

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

pka = -log ka

pka = -log 1.8*10^-5

pka =4.74

lets use the pka value in the above equation

pH=4.74+ log ([0.2]/[0.6])

pH= 4.74+(-0.477)

pH=4.26

2. Calculate the pH after 0.10 mol of HCl has been added to the original solution. Assume no volume change on addition of HCl.

Solution :-

Lets first calculate the initial moles of the HC2H3O2 and NaC2H3O2

Moles = molarity * volume in liter

Moles of HC2H3O2 = 0.60 mol per L * 1 L = 0.60 mol

Moles of NaC2H3O2 = 0.20 mol per L * 1 L =0.20 mol

After adding 0.10 mol HCl then it will react with conjugate base and forms the HC2H3O2 acid

So moles of the HC2H3O2 will increase by 0.10 mol and moles of NaC2H3O2 will decrease by 0.10 mol

So new moles are as follows

HC2H3O2 = 0.60 mol + 0.10 mol = 0.70 mol

NaC2H3O2 = 0.20 mol – 0.10 mol = 0.10 mol

Since volume is 1 L the moles is nothing but the molarity

So lets now calculate the pH using the Henderson equation

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

pH = 4.74 + log [0.10/0.70]

pH=4.74+(-0.845)

pH= 3.89

3. Calculate the pH after 0.20 mol of NaOH has been added to the original buffer solution. Assume no volume change on addition of NaOH

Solution :-

After adding 0.20 mol NaOH it will react with acid to form conjugate base

Therefore new moles of the acid and base are as follows

HC2H3O2 = 0.60 mol -0.20 mol = 0.40 mol

NaC2H3O2 = 0.20 mol + 0.20 mol = 0.40 mol

Now lets calculate the pH

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

pH= 4.74 + log [0.40 /0.40]

pH= 4.74 + 0.0

pH= 4.74