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what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl?
Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C6H5COOH) and 0.20 M sodium benzoate (C6H5COONa), (Ka (C6H5COOH) = 6.5 x 10-5)
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- hcbiochemLv 72 years ago
pH = pKa + log [benzoate]/[benzoic aicd]
pKa = - log (6.5X10^-5) = 4.19
pH = 4.19 + log (0.20 / 0.250) = 4.09
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