Question

How much sodium acetate should be added to 0.1 mole sol. Of ch3cooh to give a sol. Of ph 5.5?

Answer 1

Consider a buffer solution containing 0.1 mole each of acetic acid and sodium acetate in a 0.1 L of solution. 0.01 mole of NaOH is gradually added to this buffer solution. What will be the new [H+] ion concentration in the resulting solution?

No.of m.moles of acetic acid= 0.1 x100=10
No.of m.moles of CH3-COONa=0.1 x100=10.
0.01 mole of NaOH (ie 0.01 x100=10 m.mole) will react with 10 m.mole of acetic acid to form 10 m.moles of sodium acetate. So finally after the addition of 0.01 mole of NaOH, the concentration of acetic acid in the Solution will be 90 m.moles and the concentration of sodium acetate will be 110 m.moles.
Hence the concentration of Acetic Acid= 0.09M.
The concentration of Sodium acetate= 0.11M.
Here we can determine only the concentration of acetic acid. It is not the [H+] concentration, because it depends not only on the concentration, but also on the degree of dissociation.
That is [H+]= C.x where C= concentration =0.09 M.
If Ka=1.8 x 10^-5, then Ka=Cx2/(1-x), if x<<<1, then (1-x)=~1.
So x=(Ka/C)^0.5.
x=(1.8 x10^-5/0.09)^0.5=0.01414
So [H+]= C.x=0.09 x 0.01414
[H+] = 0.001273 =1.273 x10^-3 M
pH= pKa+ log(salt/acid)

=4.744+ log(0.11/0.09)=4.744+0.0871=4.8311

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