Question #742ba

1 Answer
Feb 14, 2018

Answer:

Unknown without data...

Explanation:

We interrogate the equilibrium....

#HOAc(aq)+H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc#

#HOAc=H_3C-C(=O)OH#

Now it is a fact that #K_a(HOAc)=10^(-4.76)=1.74xx10^-5#..

And so we set up the equilibrium expression...

#1.74xx10^(-5)=([H_3O^+][""^(-)OAc])/([HOAc])#

And if we put #[H_3O^+]=x#...then by stoichiometry...

#1.74xx10^(-5)=(x^2)/(0.0015-x)#..

And if #x"<<"0.0015#...then...#x~=sqrt(1.74xx10^-5xx0.0015)#

#x_1=1.60xx10^-4*mol*L^-1#...

And we can use this value for a second approximation....

#x_2=1.53xx10^-4*mol*L^-1#...

#x_3=1.53xx10^-4*mol*L^-1#...

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

And so #"% ionization"="moles of anion"/"moles of starting acid"xx100%#

#=(1.53xx10^-4*mol*L^-1)/(0.0015*mol*L^-1)xx100%=10.2%#

This #"% ionization"# is GREATER than for a more concentrated solution of weak acid .... at increasing dilution #"% ionization"# becomes greater for weak acids....