Find the pH of a 0.350 M aqueous benzoic acid solution. For benzoic add. Ka = 6.5 x 10^-5. a) 4.64
b) 4.19
c) 2.32
d) 11.68

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

a galvanic cell( aka voltaic cell) and electrolytic cell

explanation:

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