Calculate the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to

Question

Calculate the volumes of 0.20 M acetic acid and 0.20 M sodium acetate needed to make 20.00 mL of a buffer having pH = 5.00. Repeat analogous, separate calculations for buffers having pH = 4.00 and pH = 6.00. Hint: set up two equations with two unknowns; the first is the Henderson-Hasselbalch equation, the second is the fact that the volumes of HA (x mL) and A? (y mL) solutions should sum to 20 mL. In the Henderson-Hasselbalch equation, use the same x mL and y mL for the ratio, since the concentrations of the HA and A? solutions are the same, and the change in the ratio is due to the different volumes.

Explanation / Answer

1)pH=5.00

Volume of acetic acid=x ml

Volume of sodium acetate=y ml

Moles of acetic acid=molarity * volume=0.20M* x ml

Moles of acetate=molarity * volume=0.20M* y ml

[A-]/[HA]= 0.20M* y ml/0.20M* x ml=y/x

Or,

[A-]/[HA]= y/x

Using Henderson-Hasselbalch equation,

pH=pka + log [base]/[acid]

or, pH=pka +log [A-]/[HA]

pH=5.00

pka for acetate/acetic acid buffer=4.76

or, pH=pka +log y/x

5.00=4.76 +log y/x

0.24= log y/x

Or, y/x=10^-0.24=1.74

y/x=1.74………(1)

Also y+x=20 ml

Y=20-x………(2)

Putting eqn 2 in 1,

20-x/x=1.74

20-x=1.74x

20=x+1.74x

X=20/2.74=7.3 ml

Y=20-7.3=12.7ml

2) pH = 4.00

Similarly ,

, pH=pka +log y/x

4.00=4.76 +log y/x

-0.76= log y/x

Or, y/x=10^-0.76=0.17

y/x=0.17………(1)

Also y+x=20 ml

Y=20-x………(2)

Putting eqn 2 in 1,

20-x/x=0.17

20-x=0.17x

20=x+0.17x

X=20/1.17=17.09ml

Y=20-17.09=2.91ml

3) pH = 6.00.

, pH=pka +log y/x

6.00=4.76 +log y/x

2.76= log y/x

Or, y/x=10^2.76=575.4

y/x=575.4………(1)

Also y+x=20 ml

Y=20-x………(2)

Putting eqn 2 in 1,

20-x/x=575.4

20-x=575.4x

20=x+575.4 x

X=20/576.4= 0.035 ml

Y=20-0.035=19.965ml

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