Find the pH of a 0.14M CH3COOH solution. Ka= 1.8x 10^-5

Expert Solution

CH3COOH + H2O -----------> CH3COO- + H3O⁺

I 0.14M 0 0

C -x +x +x

E 0.14-x +x +x

Ka = [CH3COO-][H3O+]/[CH3COOH]

1.8*10^-5 = x*x/0.14-x

1.8*10^-5 *(0.14-x) = x²

x = 0.00157M

[H3O+] = x = 0.00157M

PH = -log[H3O+]

= -log0.00157

= 2.8041

queen_honey_blossom answered 1 year ago

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