posted by sparkle on .
A solution of C6H5CO2Na hydrolyzes according to the equation. If the [C6H5CO2H] in the solution after hydrolysis is 0.00000294 M, calculate the equilibrium concentration of C6H5CO2Na.
C6H5CO2-(aq) + H2O(l) = OH-(aq) + C6H5CO2H(aq)
What does Kh stand for? Is it the same as Kw?
What does Kh stand for? The clue is in the question - hydrolysis.
before you start to answer this type of question it is useful to look at any quilibrium constant that is given. In this case Kh is small and so the equilibrium lies on the reactant side. We are calculating the concentration of a reactant so expect a largish number.
Another thing to notice is that there are no units for Kh. from this we can say that [H2O] is part of the equilibrium expression as other wise Kh will have units.
The equilibrium concentration of C6H5CO2Na will be the same as C6H5CO2-
Thus we can write
for each mole of C6H5CO2H formed a mole of OH- is formed so [OH-]=[C6H5CO2H]=0.00000294 M
If the equilibrium conc [C6H5CO2-] =x then at equilibrium [H2O]=x
Kh=[0.00000294 M][0.00000294 M]/[x][x] = 1.59 x 10^-10
solve for x
I got 0.233 M for [C6H5CO2Na]
but check the maths