Question

Calculate the pH at the equivalence point in titrating 0.120 M solutions of each of the following with 8.0×10⁻²M NaOH.

...

sodium hydrogen chromate (NaHCrO4)

 

Answer 1

First assume that the volume of sodium hydrogen chromate (NaHCrO4)

, which will be analyze is 10 ml.

Given that molarity of NaHCrO4= 0.120 M

And molarity of NaOH = 8.0×10⁻²M

AND

Number of moles = molarity * volume in L

NaHCrO4= 0.120 M *10/1000

= 0.0012 moles

Since this completely dissociates, we will also have 0.0012 mol HCrO4-.

At the equivalence point, we will have neutralized all of the HCrO4-, so we will have 0 mol HCrO4- but 0.0012 mol CrO4 2- (the conjugate base) will have formed.

We also need to figure out how much NaOH was added to get to this point in order to calculate the molarity of CrO4 2-.

For NaOH: MV=moles; V=moles/M; V=0.0012 /8.0×10⁻²M

=0.015 L

Or 15 ml

So, we now have a total of .10++15=25 ml or 0.025L of solution

(the 0.010 comes from the original amount of NaHCrO4 solution was in the beaker).

Thus, the molarity of CrO4 2- = NUMBER OF MOLES /VOLUME IN L

=

0.0012/0.025=0.048 M.

The ka of the hydrogen chromate ion = 3.0*10^-7,

Now calculate the value of kb as follows:

kb=kw/ka=(10^-14)/(3.0*10^-7)=3.3*10^-8

The equilibrium molarity of CrO4 2- is 0.044-x, and the equilibrium molarities of OH- and HCrO4- will be x. So we solve:

3.3*10^-8 = x^2/(0.048-x)

Due to small value of Kb we can write 0.048-x= 0.048

Then;

3.3*10^-8 = x^2/(0.048)

x=3.98*10^-5=[OH-]

pOH=-log[OH-]=4.40

and pH +pOH = 14
pH=14-pOH
pH =14-4.40

=9.60