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3241004551 [841]
3 years ago
5

Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic add. Ka = 6.5 x 10^-5.

Chemistry
1 answer:
slava [35]3 years ago
7 0

Answer:

correct option is (a)

The solution would be using this: C6H5COOH = H+ + C6H5COO Ka = 6.5 x 10^-5 = (H+)(C6H5COO-) over

(C6H5COOH)

Let X = moles per liter (H+) and also = moles per liter (C6H5COO-)

Ka = 6.5 x 10^-5 = (X)(X) over .350 molar = acid solution 6.5 x 10^-5 = X^2 over .350

X^2 = 6.5 x 10^-5 times .350 which = 2.275 x 10^-5

x = V2.275 x 10^-5

X = 1.5083 x 10^-5 moles per liter H+

pH = -log(H+) = -log 1.5083 x 10^-5 which

= 4.6215

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