A chemist titrates 150.0 mL of a 0.2653 M carbonic acid (H2CO3) solution with 0.2196 M NaOH solution at 25 °C. Calculate the pH at equivalence. The pKa of carbonic acid is 3.60

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

the law of conservation of mass states that in a chemical reaction, the total mass of reactants is equal to the total mass of products. because atoms are only rearranged in a chemical reaction, there must be the same number of sodium atoms and chlorine atoms in both the reactants and products

wher is the rest of the question