Results for 'The hypochlorate ions OCl- has a kb=3.6x10-7. a solution is prepared by dissolving 12.0g of NAOCl (mm=74.45g/mol) in enough water to make 835 ml solution. What is the ph of the solution?

1 Answer
anor277
Apr 20, 2018

We address the equilibrium:

#underbrace(ClO^(-))_"hypochlorite ion" + H_2O(l) rightleftharpoons ClOH(aq) + HO^(-)#

We get #pH=10.42#

Explanation:

We gots #[""^(-)OCl]=((12.0*g)/(74.45*g*mol^-1))/(835xx10^-3*L)=0.193*mol*L^-1#...initially..and this undergoes some reaction with the water solvent such that...

#([HO^-][ClOH])/([ClO^(-)])=3.60xx10^-7#...we put #x=[HO^-]#...and so...

#3.60xx10^-7=(x^2)/(0.193*mol*L^-1-x)#...and so....

#x=sqrt(3.60xx10^-7xx(0.193-x))#...and in the usual way, while this is solvable EXACTLY, we make the reasonable approx. that #0.193">>"x#...and so ...

#x_1=sqrt(3.60xx10^-7xx(0.193))=2.64xx10^-4#...and we plug the given value in again to give a second approx...

#x_2=sqrt(3.60xx10^-7xx(0.193-2.64xx10^-4))=2.63xx10^-4#...and given the convergence of the approximations, this is pretty close to the true value.

But #x=[HO^-]=2.63xx10^-4*mol*L^-1#. #pOH=-log_10{2.63xx10^-4}=3.58#...and #pH=14-pOH=10.42#

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