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How much will the pH change if we add 0.15 mol NaOH to 1.00 L of a buffer that contains 1.00 mol HC2H3O2 and 1.00 mol NaC2H3O2? In case of decreasing pH the answer should be negative.

Respuesta :

kendrich

Answer:

See explaination

Explanation:

first calculate initial pH

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

[NaC2H3O2] = [HC2H3O2] = 1.0 M

log [NaC2H3O2]/[HC2H3O2] = 0

pKa = 4.74 standard value

pH = 4.74

after addition of NaOH folowing reaction takes place.

HC2H3O2 + NaOH -------------> NaC2H3O2

[NaC2H3O2] = 1.0 + 0.15 = 1.15 M

[HC2H3O2] = 1.0 - 0.15 = 0.85 M.

pH = pKa + log [NaC2H3O2] / [HC2H3O2]

pH = 4.74 + log [1.15] / [0.85]

pH = 4.87

change in pH = 4.87 - 4.74 = 0.13

change in pH = 0.13

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