Answer:
A) pH = 8.11
B) pH = 7.96
Explanation:
The pH of buffer is calculated using Henderson Hassalbalch's equatio, which is
[tex]pH=pKa+log\frac{[salt]}{[acid]}[/tex]
pKa = -logKa = -log(3.0 ✕ 10⁻⁸)
pKa = 7.5
[salt] = 0.667
[Acid] = 0.214
[tex]pH = 7.5 + log\frac{0.667}{0.214}=7.99[/tex]
A) If the solution is halved it means the concentration of both salt and acid will be the same.
The volume will change
Number of moles of acid = molarity X volume = 0.214 X 25 = 5.35 mmol
Number of moles of salt = molarity X volume = 0.667 X 25 = 16.68 mmol
moles of NaOH added = molarity X volume = 0.100 X 10 = 1 mmol
These moles of NaOH will react with acid to give same amount of salt and thus the moles of acid will decrease
New moles of acid = 5.35 - 1 = 4.35
New moles of salt = 16.68 + 1 = 17.68
the new pH will be
[tex]pH = 7.5 + log\frac{17.68}{4.35}=8.11[/tex]
B)
moles of HCl added = molarity X volume = 0.100 X 10 = 1 mmol
These moles of HCl will react with salt to give same amount of acid and thus the moles of salt will decrease
New moles of acid = 5.35 + 1 = 6.35
New moles of salt = 16.68 - 1 = 15.68
the new pH will be
[tex]pH = 7.5 + log\frac{15.68}{5.35}=7.97[/tex]
