Write the equation.
Set up an ICE chart.
I think from just a quick look that the acid and base exactly neutralize each other; therefore, the pH will be that of the salt.
Write the hydrolysis equation and set up an ICE chart and go from there.
Phenol (abbreviated HOPh) is a weak acid with a Ka= 1.29 X 10^-10. What is the pH of a mixture made by mixing 100.0 ml of a 0.700 M solution of phenol with 70.0 ml of a 1.00 M solution of NaOH?
Help please and please show work... I'm so confused.
4 answers
HOPh + OH- > H2O + OPh-
(0.1L)(0.700 mol/L)= 0.07 mol HOPh
(0.07L)(1.00 mol/L)= 0.07 mol NaOH
I 0.07 0.07 0
C -0.07 -0.07 +0.07
E 0 0 0.07
(H3O+)= mol HOPh/mol OPh- X Ka= (0 mol/0.07 mol) X 1.29 X 10^-10
I'm stuck here because the 0 makes the problem unsolvable.
(0.1L)(0.700 mol/L)= 0.07 mol HOPh
(0.07L)(1.00 mol/L)= 0.07 mol NaOH
I 0.07 0.07 0
C -0.07 -0.07 +0.07
E 0 0 0.07
(H3O+)= mol HOPh/mol OPh- X Ka= (0 mol/0.07 mol) X 1.29 X 10^-10
I'm stuck here because the 0 makes the problem unsolvable.
Oh wait I got an answer is it pH= 8.86?
right.
That's why I put in the second part in my response. The pH is the pH of the salt.
NaOPh is the Na^+ and the OPh^-. The OPh^- is hydrolyzed.
OPh^- + HOH ==> HOPh + OH^-
Kb = Kw/Ka = (HOPh)(OH^-)/OPh^-
You know Kw, Ka, and OPh^-. (HOPh)=(OH^-). Let x = each of them and solve for x. Then change to pOH and pH from there. (I didn't write the ICE table but it might make it easier for you to understand if you did that.)
That's why I put in the second part in my response. The pH is the pH of the salt.
NaOPh is the Na^+ and the OPh^-. The OPh^- is hydrolyzed.
OPh^- + HOH ==> HOPh + OH^-
Kb = Kw/Ka = (HOPh)(OH^-)/OPh^-
You know Kw, Ka, and OPh^-. (HOPh)=(OH^-). Let x = each of them and solve for x. Then change to pOH and pH from there. (I didn't write the ICE table but it might make it easier for you to understand if you did that.)