70% c Search Part 1 You obtain a buffer that o 451 moles of chlorous acid (HCIO2

Question

70% c Search Part 1 You obtain a buffer that o 451 moles of chlorous acid (HCIO2, K h10 x 10 and 0.451 moles of potassium chlorite (KCIO2 dissolved in 1.00 Lof solution. What is the pH of the buffer? 1.9S Part 2: To the buffer in part 1, you add o.122 moles of potassium hydroxide (KoH). Write the reaction of the buffer with the base. After, complete an ICE table calculating the END moles of each substance after the reaction. Hint consider potassium hydroxide to react COMPLETELY. Which of the following represents the END moles of HCIO2and Clo2 after reaction? CA HCIOz: o329 moles CIO2 0.329 moles B HCIO2:0.573 moles CIOz :0.329 moles CIO2 0.573 moles DO HCIOz: 0.573 moles CIO2 0.573 moles Part 3:

Explanation / Answer

Part 1)

We have taken 0.451 moles of HA and 0451 moles of A- in one L solution.

the Ka of the acid = 1.1 x10-2

the pH of buffer is calculated using Hendersen equation

pH = pKa + log [conjugate base] /[acid]

= (2 - log1.1) + log 0.451/0.451

= 1.9586

part 2

To this buffer 0.122 moles of KOH is added

HA + OH- ----------------> A- + H2O

0.451 0 0.451 - initial moles

- 0.122 - - change

0.329 - 0.573 - after reaction

[HClO2] = 0.329 M and [ClO2-] = 0.573 M

part 3

The pH of the new buffer after adding KOH is also calculated using Hendersen equation

pH = 1.9586 + log 0.573/0.329

= 2.1995

part 4

The moles of KOH to overwhelm the buffer, that is it can no more act as a buffer.

If we completely react all the acid with KOH, the solution is no more buffer, as it now contains only the conjugate base.

HA + OH- ----------------> A- + H2O

0.451 0 0.451 - initial moles

- 0.451 - - change

0 0 0.902 - after reaction

Now the solution has no weak acid, but only salt . So the buffer is overwhelmed by adding 0.451 moles of KOH

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