You obtain a buffer that 0.437 moles of chlorous acid (HClO2, Ka = 1.10 x 10-2)

Question

You obtain a buffer that 0.437 moles of chlorous acid (HClO2, Ka = 1.10 x 10-2) and 0.437 moles of potassium chlorite (KClO2) dissolved in 1.00 L of solution. The ph of the buffer is 1.96. You add 0.191 moles of potassium hydroxide (KOH). Potassium hydroxide to reacts COMPLETELY. Which of the following represents the END moles of HClO2 and ClO2- after reaction?

A HClO2: 0.246 moles

ClO2-: 0.246 moles

B HClO2: 0.246 moles

ClO2-: 0.628 moles

C HClO2: 0.628 moles

ClO2-: 0.246 moles

D HClO2: 0.628 moles

ClO2-: 0.628 moles

Explanation / Answer

Since, KOH being a strong base reacts with as much equivalent moles of acid as it has to form 0.191 moles KClO2.

So, end moles of HClO2 = ( 0.437 - 0.191) moles

=0.246 moles

And end moles of ClO2- = (0.191 + 0.437) moles

=0.628 moles

So correct choice is B.

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