A buffer is made by adding 0.300 mol CH3COOH nd 0.300 mol CH3COONa to enough water to make 1.000 M of solution. The pH of the buffer is 4.7?

(A)
In the question, "to make 1.000 M of solution" should be "to make 1.000 L of solution" instead.

Consider the dissociation of CH₃COOH:
CH₃COOH(aq) + H₂O(ℓ) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) ….. pKa

In the 1.000 L of the original solution:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4.74 = pKa + log(0.300/0.300)
pKa = 4.74

After addition of 0.02 mol HCl:
CH₃COO⁻(aq) + H₃O⁺(aq) → CH₃COOH (aq) + H₂O(ℓ)
[CH₃COO⁻]/[CH₃COOH] = (0.300 - 0.020)/(0.300 + 0.020) = 0.280/0.320

Consider the dissociation of CH₃COOH:
CH₃COOH(aq) + H₂O(ℓ) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)

Apply Henderson-Hasselbalch equation again:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
pH = 4.74 + log(0.280/0.320)
pKa = 4.68

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(B)
When 0.020 mol HCl is added to 1.000 L pure water,
[H₃O⁺] = (0.020 mol) / (1.000 L) = 0.020 M

pH = -log[H₃O⁺] = -log(0.020) = 1.70