Question

a buffer solution contains 0.462 M KHCO3 and 0.260 M K2CO3. Determine the pH change when 0.052 mol HBr is added to 1.00 L of the buffer

Answer 1

step I:

pOH of buffer solution

To calculate the pOH of the buffer , use the Henderson - Hasselbalch equation:
pOH = pKb + log ( [salt]/[base])
pOH = 10.25 + log (0.462/0.260)
pOH = 10.25 + log 1.78
pOH = 10.25 + 0.250
pOH = 10.5

Step II:

pH of buffer solution

pH +pOH =14
pH = 14- 10.5

pH = 3.5

Step III:
Now you add 0.052 mol HBr to the 1000mL or 1.00 L of buffer.

HBr reacts with K2CO3 in 1:1 molar ratio to produce; salt KHCO3, the chemical reaction of this conversion is following:

HBr + K2CO3 → KHCO3 + KBr.


This will react with the K2CO3 in solution and produce same number 0.052 mol KHCO3 and the mol of K2CO3 will be reduced by 0.052 mol

Mol K2CO3 in 1000 mL or 1.00 L of 0.260 M solution = 1000/1000*0.260 = 0.260 mol K2CO3
Mol KHCO3 in 1000 mL or 1.00 L of 0.462 M solution = 1000/1000*0.462 = 0.462 mol KHCO3

After adding the0.052 mol HBr to the reaction which produce 0.052 mol KHCO3, then
mole of K2CO3 = 0.260-0.052 = 0.208 mol K2CO3

Mol KHCO3:

0.462 + 0.052= 0.514 mol KHCO3.

Step IV:

Again calculated the pOH of the solution with new moles:
pOH = pKa + log ([salt] / [base])
pOH = 10.25 + log (0.514/0.208)
pOH = 10.25 + log 2.47
pOH = 10.25 + 0.39
pOH = 10.64

Step V:

Now new pH of this solution is following:
pH = 14.00 – 10.64
pH = 3.36

Step VI:

Change is pH is calculated as follows:

Before addinf HBr ;pH = 3.5 – after adding HBr ; pH 3.36=0.14

Hence the pH of change when 0.052 mol HBr is added to 1.00 L of the buffer is 0.14.