Cn some1 plz help me

Question

1.44g of sodium benzoit is added to 100mL of the 0.150M benzoic acid solution.calculate the pH of the resulting buffer solution assuming the volume of the solution does not change on the addition of the solid.

Answer 1

my text has K= 6.3e-5 for benzoic acid

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find moles of sodium benzoate using its molar mass:

1.44 grams @ 144.11 g/ mol = 0.00999 moles of sodium benzoate

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benzoic acid --> H+  & benzoate-

K = [H+] [benzoate-] / [benzoic acid]

6.3e-5  = [H+] [0.999] / [0.150]

H+ = 9.46e-6 molar

pH = 5.024

your answer is(2 sig figs): pH = 5.02

(the K only had 2 sig figs, & I am assuming that the 100 mls likely may have had 3)