Chemistry, 25.07.2019 07:00 mazolethrin6876
Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5. a. 2.29 b. 9.81 c. 5.11 d. 8.89 e. 11.71
Answers: 1
![Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X} (assume: 0,40-X\approx0,40)\\ \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\ \\ pH=-log([H_3O^{+}]=2,29](/tpl/images/0130/2057/cea00.png)
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Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic ac...
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