Answer:
5.90
Explanation:
Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol
Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol
CH3COO- + HCl => CH3COOH + Cl-
Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol
Moles of CH3COOH formed = moles of HCl added = 0.0005 mol
pH = pKa + log([CH3COO-]/[CH3COOH])
= -log Ka + log(moles of CH3COO-/moles of CH3COOH)
= -log(1.78 x 10^(-5)) + log(0.007/0.0005)
= 5.90
Answer:
The correct answer is 5.895.
Explanation:
The reaction will be,
CHCOO⁻ + H+ ⇔ CH₃COOH
Both the HCl and the acetate are having one n factor.
The millimoles of CH₃COO⁻ is,
= Volume in ml × molarity = 10 × 0.75 = 7.5
The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5
Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0
The volume of the solution is, 10+5 = 15 ml
The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15
The molarity of CH₃COOH is 0.5/15
pH = pKa + log[CH₃COO⁻]/[CH₃COOH]
= 4.74957 + 1.146
= 5.895
