Fe(s) + CuSO4(aq) ⇒ Cu(s) + FeSO4(aq)
1 answer:
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Answer:
A) pH = 8.11
B) pH = 7.96
Explanation:
The pH of buffer is calculated using Henderson Hassalbalch's equatio, which is
pKa = -logKa = -log(3.0 ✕ 10⁻⁸)
pKa = 7.5
[salt] = 0.667
[Acid] = 0.214
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
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