Answer:
9.3
Explanation:
This is long and complicated so get ready
We are going to use the conjugate base of carbonic acid with water to make carbonic acid and OH- (Na is simply a spectator ion and is irrelavent here)
Let the conjugate base be A- and Carbonic acid be HA
A- + H20 ⇄ HA + OH-
To find the concentration of A- we must find the concentration of the reactants given. We know this will be equal because it is a strong base and all of it disassociates.
to get moles of acid we take the concentration and multiply by liters to cancel
.2653 x .150 = .039795 mol HA
Because it is at equivalence point we know the moles will be equal. We are given the concentration so we only have to solve for liters
We plug it into the equation and found: .181 L
Now use moles and combined volums to fins concentrarion which is .120 M
Now plug that use the Ka converted to Kb to find the cincentrations of HA and OH-
Ka is (10^-3.60) = 2.4E-4
Kb x Ka is 10^-14
Kb = 3.98E-11
Now we know Kb = [HA] [OH] / [A-]
Solve for this through algebra by using x for the values you dont know
youll find x^2 = 3.3E-10
X = 1.8 E -5
this is the OH- concentration
-log [oh] = pOH
pOH = 4.73
We know 14-pOH = ph so pH= 9.3
