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How many moles of HIO2 and its conjugate base (IO2−) are needed to create 0.45 L...

How many moles of HIO2 and its conjugate base (IO2−) are needed to create 0.45 L of a 1.00 M HIO2 buffer solution with a pH of 4.69? (The Ka of HIO2 is 3.2 ✕ 10−5.)

HIO2?

IO2-?

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Answer #1

pH = pKa + log[A-/HA]

[A-] = IO2-; [HA] = HIO2

pH = 4.69; pKa =4.49

4.69 = 4.49 + log[A-/HA]

0.2 = [A-]/[HA]

[A-] = 0.2[HA] (equation 1)

The total molarity of the buffer system is 1.0 M, so:

[A-] + [HA] = 1.0M (equation 2)

Plug equation 1 into equation 2 to get:

0.2[HA] + [HA] = 1.0

1.2[HA] = 1.0M

[HA] = 0.833M
[A-] = 1.2[HA] = 1 M

So the amount of HIO2 would be:
0.833 M * 0.45 L = 0.374 moles

So the amount of IO2- would be:
1 M * 0.45 L = 0.45 moles

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