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What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5 M CHOOH...

What volume of 2.0 M NaOH must be added to 100.0 mL of 2.5 M CHOOH (Ka= 1.8 x10-4) to prepare a buffer with a pH of 4.0?

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Homework Answers

Answer #1

This is a reaction between strong base and weak acid so complete formation of HCOONa is expected

When Ka = 1.8 x 10-4 then pKa will be log (Ka)

pKa = -log (Ka)

pKa = -log (1.8 x 10-4)

pKa = 3.744

moles of HCOOH = (100 mL x 2.5M)/1000 mL = 0.25 moles

Reacts with NaOH x moles to give HCOO- = x

HCOOH remaining is 0.25 -x

Henderson-Hasselbach equation

pH = pKa + log [A-]/[[HA]

4.0 = 3.744 + log x/0.25-x

4.0 - 3.744 = log x/0.25-x

0.256= log x/0.25-x

x/0.25-x = 100.256

x/0.25-x = 1.80

x = 1.80(0.25-x)

x = 0.45 - 1.8x

2.8 x = 0.45

x = 0.1609 This is the moles of NaOH

Since the NaOH concentration is 2.0 M the volume would be

(2 x volume)/1000 = 0.1609

2 x volume = 160.9

volume = 160.9/2

volume = 80.49 mL of 2.0 M NaOH is needed

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