Question

What will be the ph value of buffer solution having 400ml of 0.5MCH3COOH and 800ml of 0.1M CH3COONa which is 1.3% ionised in solution?

Answer 1

Recall,

`I = ([H^+] ) / [HA]_"initial"*100`

Let's start by calculating the concentration of protons at equilibrium,

`[H^+]_"eq" = (I*[HA]_"initial")/100 approx 5.42*10^-5"M"`

Hence,

`"pH" = -log[H^+] approx 4.27`

This is reasonable, because the `"pK"_"a" approx 4.76`, and there is more acid present in this case. Given no other data, we can approximate the pH above.

Answer 2

pH = 3.7

Explanation:

Ethanoic acid is a weak acid:

`sf(CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)`

`sf(K_a=([CH_3COO^-][H^+])/([CH_3COOH]))`

The % ionised enables us to calculate `sf(K_a)`

The degree of dissociation `sf(alpha)` is therefore 0.013.

We construct an ICE table based on mol/l. Let C be the initial concentration of the acid:

`sf(color(white)(xxx)CH_3COOHrightleftharpoonsCH_3COO^(-)+H^+)`

`sf(Icolor(white)(xxxx)Ccolor(white)(xxxxxxxxxxxx)0color(white)(xxxxx)0)`

`sf(Ccolor(white)(xx)-Calphacolor(white)(xxxxxxxxx)+Calphacolor(white)(xxx)+Calpha)`

`sf(Ecolor(white)(xx)C-Calphacolor(white)(xxxxxxxxxx)Calphacolor(white)(xxxx)Calpha)`

`:.` `sf(K_a=(C^2alpha^2)/(C(1-alpha)))`

`sf(K_a=(Calpha^2)/((1-alpha))`

`sf(K_a=(0.5xx0.013^2)/((1-0.013))=8.56xx10^(-5))`

Rearranging the mass action expression for `sf([H^+])`:

`sf([H^+]=K_axx([CH_3COOH])/([CH_3COO^-]))`

Theses are equilibrium concentrations. For `sf(K_a)` values between `sf(10^-4)` to `sf(10^(-9)`we can assume that they approximate to initial concentrations.

Since the total volume is common I will just use moles:

`sf(n_(CH_3COOH)=cxxv=0.5xx400/1000=0.2)`

`sf(n_(CH_3COO^-)=cxxv=0.1xx800/1000=0.08)`

`:.` `sf([H^+]=8.56xx10^(-5)xx0.2/0.08=2.14xx10^(-4)color(white)(x)"mol/l")`

`sf(pH=-log[H^+]=-log[2.14xx10^(-4)]=3.7)`