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.
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.
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.
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.)
To find the pH of the mixture, you need to consider the reaction between phenol (HOPh) and NaOH. Phenol is a weak acid, and NaOH is a strong base, so they will react to form a salt, sodium phenolate (NaOPh) and water (H2O).
First, let's calculate the number of moles of phenol and NaOH in the given solutions:
Number of moles of phenol:
Moles = (concentration in M) x (volume in L)
Moles of phenol = (0.700 M) x (0.100 L) = 0.070 moles
Number of moles of NaOH:
Moles = (concentration in M) x (volume in L)
Moles of NaOH = (1.00 M) x (0.070 L) = 0.070 moles
Notice that the number of moles of phenol and NaOH is the same, indicating that they will react in a 1:1 ratio.
Now, let's calculate the concentration of remaining phenol and NaOH after the reaction:
Moles of phenol remaining = 0.070 - 0.070 = 0 moles
Moles of NaOH remaining = 0.070 - 0.070 = 0 moles
Since all of the phenol is consumed in the reaction, we can ignore its contribution to pH. The remaining solution is now a solution of NaOH, so we need to calculate the concentration of NaOH:
Concentration of NaOH = (moles of NaOH remaining) / (volume in L)
Concentration of NaOH = 0 moles / (0.100 L + 0.070 L) = 0 M
At this point, we have calculated that the concentration of NaOH is 0 M. This means that there are no hydroxide ions (OH-) in the solution, and therefore the pH is not influenced by the presence of the NaOH solution.
Since the pH is solely determined by the remaining solution of phenol, we can calculate the pOH of the solution and convert it to pH using the relation: pH = 14 - pOH.
Now, let's calculate the pOH:
pOH = -log10 (OH- concentration)
Since we have already determined the concentration of NaOH to be 0 M, the concentration of OH- in the solution is also 0 M.
pOH = -log10 (0) = undefined
We cannot calculate the pOH and therefore the pH since there are no hydroxide ions present in the solution.