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

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

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?
 
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Discussion

Borek said:
Show how you got this result.

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
 
Last edited:
daivinhtran said:
.090476 M .00476M
-x +x +x
.090476 M x .00476M

Check the last line again.
 
Borek said:
Check the last line again.

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.
 
Borek said:
Check the last line again.

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