Calculate the ph of a solution prepared by mixing 15.0 ml of 0.10 m naoh and 30.0 ml of a 0.10 m benzoic acid solution. (benzoic acid is monprotic; its dissociation constant is 6.5 x 10^-
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Ответ:
Moles of NaOH = (0.10 mol/L) (0.015 L) = 0.0015 mol
Moles of benzoic acid = (0.100 mol/L) (0.0300 L) = 0.003 mol
NaOH and benzoic acid react in a 1:1 molar ratio. This means they are a buffer.
After the reaction
moles of sodium benzoate = 0.00150 mol
moles of benzoic acid = 0.00150 mol
pH = pKa + log [base / acid]
pH = -log(6.5 x 10^-5)
pH = 4.187
pH = 4.19
Ответ: