Question

How many mL of 0.05 M sodium acetate should be added to 100 mL of 0.05M acetic acid to make a buffer of pH 5.1? What is the molarity of the resulting buffer with respect to acetate (Acetate + Acetic Acid)? pKa acetic = 4.76

Answer 1

Step-by-step explanation:

Let us assume that volume of acetic acid added is V ml.

So,
`[CH_(3)COOH] = (0.05 * 100)/(100 + V)`

and,
`[CH_(3)COONa] = (0.05 * V)/(100 + V)`

Expression for the buffer solution is as follows.

pH =
`pK_(a) + log ([CH_(3)COONa])/([CH_(3)COOH])`

5.1 =
`4.76 + log (0.05 * V)/(0.05 * 100)`

0.34 = log V - 2

log V = 2.34

or, V = 218.77 ml

Now, we will calculate the molarity of the buffer with respect to acetate as follows.

=
`[CH_(3)COO^(-)] + [CH_(3)COOH]`

=
`(0.05 * 218.77)/(318.77) + (0.05 * 100)/(318.77)`

= 0.0499 M

or, = 0.05 M (approx)

Thus, we can conclude that molarity of the resulting buffer with respect to acetate is 0.05 M.