19) The Ka of benzoic acid is 6.30 × 10-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium benzoate and 50.0 mL of 1.00 M benzoic acid is __.
A) 1.705
B) 0.851
C) 3.406
D) 4.201
E) 2.383
3 Answers
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pKa = -log(6.30e-5)
pKa = 4.20
0.050L * 1.00M = 0.050mols benzoic acid
0.050L * 1.00M = 0.050mols salt
0.050mols / 0.1L = 0.50M
0.050mols / 0.1 = 0.50M
pH = 4.20 + log(0.50/0.50)
pH = 4.20
Choice D
Source(s): Live4 -
use the H-H equation for first area. you may extremely sparkling up for the ratio Then sparkling up for the A-, it extremely is likewise the quantity of salt that replaced into created and is likewise the quantity of base that replaced into further. Convert the molarity to moles
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moles benzoic acid = 0.05 L x 1 mol/L = 0.05 moles
moles benzoate = 0.05 L x 1 mol/L = 0.05 moles
total volume = 50 ml + 50 ml = 100 ml = 0.1 L
[benzoate] = 0.05 mol/0.1 L = 0.5 M
[benzoic acid] = 0.05 mol/0.1 L = 0.05 M
pH = pKa + log [salt]/[acid]
pKa = -log Ka = 4.2
pH = 4.2 + log 0.5/0.5
pH = 4.2
