Answer:
pH = 4.58
Explanation:
The reaction of NaOH with acetic acid, CH₃COOH occurs as follows:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
<em>Moles that react:</em>
NaOH = 10mL = 0.010L * (0,240mol / L) = 0.0024 moles NaOH
CH₃COOH = 50.0mL = 0.050L * (0.120mol / L) = 0.0060 moles CH₃COOH
That means after the reaction you will have:
CH₃COOH: 0.0060 mol - 0.0024 mol = 0.0036 moles
CH₃COO⁻Na⁺: 0.0024 moles
in solution, you will have the mixture of a weak acid (Acetic acid), with its conjugate base (sodium acetate, CH₃COO⁻Na⁺). And pH of this buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
For Acetic buffer pKa = 4.76:
pH = 4.76 + log [CH₃COO⁻Na⁺] / [CH₃COOH]
<em>Where [] is molarity of each species or moles</em>
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Replacing:
pH = 4.76 + log [0.0024 moles] / [0.0036 moles]
<h3>pH = 4.58</h3>
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