For the next three problems, consider 1.0 L of a solution which is 0.6 M HC 2 H
ID: 885564 • Letter: F
Question
For the next three problems, consider 1.0 L of a solution which is 0.6 M HC2H3O2 and 0.2 M NaC2H3O2 (Ka 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.
Explanation / Answer
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