Ignore activity effects for the following questions:
A buffer is prepared by adding 25.00 mL of 0.100 M NaOH to 0.2525 g of benzoic acid, and diluting to 100.0 mL. What is the pH?
Mass of benzoic acid = 0.2525 g
Mass of NaOH = 0.100 M
We know molar mass of benzoic acid = 122.12 g / mol
The moles of benzoic acid = 0.2525 / 122.12
= 0.00206
moles of NaOH = 0.100 x 25 / 1000
= 0.0025
after combining both we get ,
moles of acid = 0.00206 - 0.0025
= - 0.02294
moles of salt = 0.0025
[acid] = - 0.02294 / 0.1
= -0.2294M
[salt] = 0.0025 / 0.1
= 0.025 M
We know , pKa of benzoic acid = 4.20
Let us consider ,
pH = pKa + log [salt] / [acid]
pH = 4.20 + log [0.025] / [-0.2294]
pH = 5.16
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