For this past lab, \"The Properties of Buffers,\" we used 15mL of acetic acid an

Question

For this past lab, "The Properties of Buffers," we used 15mL of acetic acid and 15mL of acetate ion to make a buffer. Both have a molarity of 0.1M.
We measured pH and got 4.39. We have to solve for pKa and use an ICE table to show how it can help us.
1) If is use pH=pKa+log(0.1/0.1), I get 4.39. Does this mean pH=pKa or am I missing something?
2) What would the ICE table look like? For this past lab, "The Properties of Buffers," we used 15mL of acetic acid and 15mL of acetate ion to make a buffer. Both have a molarity of 0.1M.
We measured pH and got 4.39. We have to solve for pKa and use an ICE table to show how it can help us.
1) If is use pH=pKa+log(0.1/0.1), I get 4.39. Does this mean pH=pKa or am I missing something?
2) What would the ICE table look like?
We measured pH and got 4.39. We have to solve for pKa and use an ICE table to show how it can help us.
1) If is use pH=pKa+log(0.1/0.1), I get 4.39. Does this mean pH=pKa or am I missing something?
2) What would the ICE table look like?

Explanation / Answer

millimoles of CH3COOH = 0.1 x 15 = 1.5

millimoles of CH3COO- = 0.1 x 15 = 1.5

pH = 4.39 given

in fact there is no need of ICE table . because buffer concentrations are clearly given

direct ICE table:

CH3COOH ----------------------------> CH3COO- + H+

1.5                                                     1.5              1.5

pH = pKa + log [CH3COO- / CH3COOH]

4.39 =pKa + log (1.5 / 1.5)

pKa = 4.39

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