Answer:
pH = 4.05
Explanation:
The pH of the benzoic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
<em>Where pKa is -logKa = 4.187</em>
pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]
<em>Where [] can be understood as moles of each specie.</em>
Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.
<em>Initial moles:</em>
<em />
Initial moles of benzoic acid and sodium benzoate are:
Acid: 250mL = 0.250L ₓ (0.250 moles / L) = <em>0.0625 moles of benzoic acid</em>
Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = <em>0.050 moles of sodium benzoate</em>
<em>Moles after reaction:</em>
Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:
HCl + C₆H₅COONa → C₆H₅COOH + NaCl
That means after reaction moles of both species are:
Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles
Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles
Replacing in H-H equation:
pH = 4.187 + log [0.0475] / [0.065]
pH = 4.05
