Question

How much HCl must be added to a liter of buffer that is 1.4 M in acetic acid and 0.80 M in sodium acetate to result in a buffer pH of 4.09?

Answer 1

CH3COOH ⇌CH3COONa

1.4 M ..... .. 0.80 M

Suppose, x M of HCl is added then,

CH3COOH ⇌ CH3COONa + H^+

1.4 M ... 0.80M + xM

1.4+x................... 0.80-x

Thus, at equilibrium,

[CH3COOH] = (1.4+x) M

[CH3COONa] = (0.80-x ) M

Now, using Henderson hesselbatch equation:

pH = pka + log { [CH3COONa]/[CH3COOH] }

where, pKa (CH3COOH) = 4.75

Now putting the values,

4.09 = 4.75 + log (0.80-x/1.4+x)

=> -0.66 = log (0.80-x/1.4+x)

=> 0.2188= 0.80-x/1.4+x

=>0.2188(1.4+x) = 0.80-x

=> 0.3063+ 0.2188x = 0.80-x

=> 0.2188x+x = 0.80-0.3063

=> 1.2188x = 0.4937

=>x = 0.4051M

Now, Molarity = No of moles / Volume in litres

=> No of moles of HCl = Molarity * Volume

=> No of moles (HCl ) = 0.4051 * 1 = 0.4051 moles

Now, No of moles = mass /molar mass

=> Mass = No of moles * Molar mass

So, Mass of HCl = 0.4051mole * 36.46g/mol

=> Mass of HCl= 14.78 g

Thus, 14.78 g of HCl must be required.

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