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?
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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