Question

You have 100ml of 1.0M PIPES buffer at the pH = 5.76. The pKa for PIPES is 6.76. How much 10 M NaOH would you need to add to change the pH from 5.76 to 6.76? Do not concern yourself with dilution effects.

a. 2.0 ml

b. 4.0 ml

c. 8.0 ml

d. 9.8 ml

The answer is B I am just unsure how to arrive there.

Answer 1

Ans: B

According to Henderson–Hasselbalch equation, pH of the buffer is given by

pH = pKa + log ([PIPES-H]/[PIPES])

So, ([PIPES-H]/[PIPES]) = 10^(pH-pKa)

where, PIPES-H = conjugate base of PIPES (deprotonated form)

            PIPES = weak acid

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Initial

Initial pH = 5.76

So, ([PIPES-H]/[PIPES]) = 10^(pH-pKa) = 10^(5.76-6.76) = 0.1

==> [PIPES-H] = 0.1 x [PIPES]

Meaning, 10% of the PIPES is deprotonated

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At pH = 6.76:

([PIPES-H]/[PIPES]) = 10^(pH-pKa) = 10^(6.76-6.76) = 1

==> [PIPES-H] = 1 x [PIPES]

Meaning, 50% of the PIPES is deprotonated

==> So, there will a change of 40% of PIPES ----> PIPES-H when the pH is changed from 5.76 to 6.76.

Reaction: 1 mol PIPES + 1 mol NaOH ---> 1 mol PIPES-H

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Millimoles of PIPES taken = Molarity x volume(mL) = 1.0 M x 100 mL = 100 mmol

40% of 100 mmol is 40 mmol.

40 mmol of PIPES demands 40 mmol of NaOH to become 40 mmol of PIPES-H.

Molarity of NaOH taken = 10 M

Volume of 10 M NaOH taken = (moles/Molarity) = 40 mmol/10M = 4 mL