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

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Chemistry

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

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

Â Â Â Â Â IÂ Â Â Â Â Â Â Â Â 0.14MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

Â Â Â Â Â CÂ Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

Â Â Â Â Â EÂ Â Â Â Â Â Â Â 0.14-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +x

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

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

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 1.8*10^-5 *(0.14-x) = x²

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â = 0.00157M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [H3O+]Â Â = xÂ Â = 0.00157M

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â PHÂ Â Â = -log[H3O+]

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = -log0.00157

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.8041

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

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

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