How do you calculate the pH of an aqueous solution of 0.0726 M potassium sulfite?

1 Answer
Jul 11, 2018

Answer:

You ain't given us any data with respect to #pK_a# of #"bisulfite"#, or #pK_b# of #SO_3^(2-)#....I gets #pH~=10#

Explanation:

This site reports that the #pK_a# of #HSO_3^(-)# is #6.97#....and therefore, the #pK_b# of #SO_3^(2-)#, its conjugate base is #14-6.97=7.03#..(because #pK_a+pK_b=14# in aqueous solution)….

And so we address the following equilibrium....

#SO_3^(2-)+H_2O(l) rightleftharpoonsHSO_3^(-) +HO^-#

And thus #K_b=10^(-7.03)=([HSO_3^(-)][HO^-])/([SO_3^(2-)])#

Now INITIALLY, #[SO_3^(2-)]=0.0726*mol*L^-1#...and if #x*mol*L^-1# sulfite ion associates then....

#K_b=10^(-7.03)=(x^2)/(0.0726-x)#...we propose that #0.0726">>"x#...and so....

#x_1=sqrt(10^(-7.03)xx0.0726)=8.23xx10^-5*mol*L^-1#...and we use this value for a successive approximation...

#x_2=sqrt(10^(-7.03)xx(0.0726-8.23xx10^-5))=8.23xx10^-5*mol*L^-1#

And since the approximations have converged, I am prepared to accept this as the true value....

But #x=[HO^-]#...#pOH=-log_10(8.23xx10^-5)=4.08#....#pH=14-pOH=14-4.08=9.92#

(Why? Because #pH+pOH=14# in aqueous solution under standard conditions...)