Answer:
The pH of the buffer is 3.90
Explanation:
The mixture of a weak acid CH3COOH and its conjugate base CH3COO produce a buffer that follows the equation:
pH = pKa + log [A-] / [HA]
<em>Where pH is the pH of the buffer, pKa is the pKa of acetic acid (4.75), and [A-] could be taken as the moles of the conjugate base and [HA] the moles of thw weak acid.</em>
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To solve this question we need to find the moles of the CH3COOH and CH3COO- after the reaction with HCl:
CH3COO- + HCl → CH3COOH + Cl-
<em>The moles of CH3COO- are its initial moles - the moles of HCl added</em>
<em>And moles of CH3COOH are its initial moles + moles HCl added</em>
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Moles CH3COO-:
Initial moles = 0.100L * (0.010mol / L) = 0.00100moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.000500 moles
Moles CH3COOH:
Initial moles = 0.100L * (0.040mol / L) = 0.00400moles
Moles HCl = 0.010L * (0.050mol / L) = 0.000500 moles
Moles CH3COO- = 0.003500 moles
pH is:
pH = 4.75 + log [0.000500] / [0.00350]
<em>pH = 3.90</em>
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<h3>The pH of the buffer is 3.90</h3>
