The pH of the base is 4.42.
No of moles of CH₃COOH = 47.0mL x L/1000L x 0.313 mol/L = 0.0147 mol
No of moles of OH⁻ added = 4.97 mL x L/1000 mL x (0.481 mol Ba(OH)₂)/L × (2 mol OH⁻)/(1 mol Ba(OH)₂) = 0.00478 mol
CH₃COOH + OH⁻ → CH₃C00⁻ + H₂O
Total volume = 47.0 mL + 4.97 mL = 51.97 mL = 0.05197 L
Concentration of CH₃COOH, [CH₃COOH] = 0.00992 mol/0.05197 L = 0.191M
Concentration of CH3C00⁻, [CH3C00⁻] = 0.00478 mol/0.05197 L = 0.0920 M
pKᵇ, of CH₃COOH = -log Kₐ = -log (1.8x10⁻⁵) = 4.74
According to Henderson equation,
pH = pKₐ + log [CH3C00⁻] / [CH₃COOH]
= 4.74 + log 0.0920 / 0.191 = 4.42
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