Buffer (pH of a weak acid decreases when we add NaOH!! Yes, really..

daivinhtran

This is not homework, but just something that I find not following general rule. Just wondering why.
So
We have the titration of 20.0 mL of .100M HF with .100M NaOH

With 0mLof NaOH, the pH=2.08 (by calculating)

But with 1.00mL of NaOH, the pH is 1.888 (by calculating)

BUT then with 2.00ml of NaOH, the pH increases to 2.213
After that, it keeps increasing and follow the rule

given the Ka= 6.8e-4

Can someone explain it?

Borek

Show how you got this result.

daivinhtran

For V of NaOH=0mL

The equilibrium equation is
HF <=> H+ + F-
.1M
-x +x +x
.1 x x

(x^2)/.1 = Ka = 6.8e-4
x= .008246 = [H+]
pH = 2.083745544

With V of Naoh=1mL

HF + OH- <=> F- + H20
(.02)(.1) (.001)(.1)
.002mol .0001mol
-.001mol -.001mol +.001mol
= .0019mol 0 .0001mol

=> [HF]=.0019mol/(.021L) = .090476 M
[F-]= .0001mol/(.021L) = .00476M

Equilibrium equation:
HF <=> H+ + F-
.090476 M .00476M
-x +x +x
.090476 M x .00476M

.00476x/.09476 = 6.8e-4
[H+] = 1.86847 = x
pH= 1.868472988

Borek

Check the last line again.

daivinhtran

Yes, I did, sir.

My teacher also checked it but she had no clue why it's like that..

SHe also did the same calculation and get the same pH.

daivinhtran

we're probably not supposed to ignore the x in this one, right?

Borek

Don't guess, try. The compare numbers you get with the numbers you got earlier.

Simple check is always to plug concentrations you get into Ka and seeing if the calculated value agrees with the given one.