Question

what is the value of PH of 2% (w÷v) na2co3 solution?​

Answer 1

Na2CO3, sodium carbonate would generate Na+ and carbonate ion. Carbonate ions should be considered a weak base, the conjugate base of bicarbonate. Some of those carbonate ions will combine with H+ ions to make bicarbonate ions, decreasing the concentration of H+ ions and raising the pH. We could write the reaction as

carbonate + H2O → bicarbonate + OH-

The dissociation constant for bicarbonate ion as an acid is Ka=4.7*10^(-11) ,

so the dissociation constant for carbonate ion as a weak base is

Kb=Kw/Ka=[bicarbonate]*[OH-]/[carbonate]=1*10^(-14)/4.7*10^(-11)=2.13*10^(-4)

The total molar carbonate concentration is

(20 g Na2CO3/1 L solution)(1 mol/138.2 g) = 0.1447 M

We expect the final molar concentration of bicarbonate to be x << 0.1447. Its formation will lead to forming an equal molar amount of OH-. We calculate assuming that the concentration of OH will also be x , and that the final concentration of carbonate ion will only be negligibly reduced so that 0.1447 - x = approximately 0.1447.

With that, substituting into [bicarbonate]*[OH-]/[carbonate] = 2.13*10^(-4) , we get

x * x / 0.1447 = 2.13*10^(-4) → x^2 = 0.1447 * 2.13*10^(-4) → x^2 = 3.08 * 10^(-5) → x = 5.55*10^(-3) .

With [OH-] = 5.55*10^(-3) , [H+] = 10^(-14) / 5.55*10^(-3) = 1.80*10^(-12) , and

pH = -log( 1.80*10^(-12) ) = 11.7 .