Solve pKa of unknown monoprotic Acid

[Comrade]

I've been trying to figure this one out for a while.

A 0.1276g sample of a monoprotic acid was dissolved in 25mL H₂O, and titrated with a .0633 M NaOH solution. Volume required to reach equivalence point was 18.4 mL.
It asked me to calculate the molar mass, which i found to be 116g/mol:
0.0633 x .0184 = .0011 moles NaOH.
1 mol NaOH = 1 mol Acid = .0011 moles Acid.
$$\frac{.1276g}{.0011}$$ = 116 g/mol

Then it asks to find the K_a of the acid after the addition of 10 mL and pH of 5.87.
I tried using the formula: pH = pK_a + log $$\frac{[A-]}{[HA]}$$
but the answer in the back of the text doesn't match up to what i come up with, so I'm guessing that isn't the right way of doing this.

 

[Borek]

Show details of your pKa calculation - approach seems to be correct.

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[Comrade]

I calculated for moles in 10mL of NaOH to be 6.3*10^-4
1 mol of NaOH = 1 mol of monoprotic acid,
moles monoprotic acid = 6.3*10^-4

then, using:
pH = pKa + log$$\frac{[A-]}{[HA]}$$
If my reasoning is correct, log$$\frac{[A-]}{[HA]}$$ will equal 1 since [A-] and [HA] will equal the moles of HA (6.3*10^-4)

solving for Ka i get:

-log Ka = 5.87
10^-5.87 = Ka = 1.35*10^-6

answer in the text is 1.6*10^-6

 

[chemisttree]

Hey Commie! Your moles of NaOH is indeed equal to the number of moles of the monoprotic acid but it isn't 6.3 X 10^-4. Do over. Remember that #moles=volumeXconcentration (of NaOH, 0.0633M and 18.4 mL).

 

[Comrade]

Alright, I just keep repeating my same thought pattern or something, because I keep coming up with:

.0633 M x .01L = .000633moles NaOH

Someone save me from my sinful ways!

edit: woah wait a minute. okay, so why am i using the 18.4 mL of base still? if it's asking for Ka after the addition of 10mL and pH 5.87

 

[chemisttree]

AGHHHH! My bad!

Yes, you are right, the number of moles of acid that have been converted to A- is 6.33 X 10^-4. What would 0.0117 - 6.3X10^-4 represent?

6.33 X 10^-4 is not [HA] in your equation... it's A- (moles). Divide the moles of A- by the volume in L to get [A-].

Remember that HA is not equal to [HA] and A- is not equal to [A-].

 

[Comrade]

A-ha!
Okay, here it is.

after addition of 10mL final volume is .035L. So 6.3*10^-4 / .035L = .018 M [A-]

.00116 moles HA - 6.3*10^-4 = 5.3*10^-4 moles (HA left in flask).
5.3*10^-4 / .035 = .015 M [HA]

then:
solving for pKA,
pKa= 5.87 - log(.018/.015)
pKa= 5.79
Ka = 10^-5.79
Ka = 1.62*10^-6

Thanks chemisttree. This problem would've been bugging me for the rest of my life.

 

[Borek]

Note: you don't need concentrations in this case, number of moles will do. Volume is the same for both nominator and denominator, so it cancels out.

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[chemisttree]

A-ha!
Okay, here it is.

after addition of 10mL final volume is .035L. So 6.3*10^-4 / .035L = .018 M [A-]

.00116 moles HA - 6.3*10^-4 = 5.3*10^-4 moles (HA left in flask).
5.3*10^-4 / .035 = .015 M [HA]

then:
solving for pKA,
pKa= 5.87 - log(.018/.015)
pKa= 5.79
Ka = 10^-5.79
Ka = 1.62*10^-6

Thanks chemisttree. This problem would've been bugging me for the rest of my life.

One minor point... significant figures. If significant figures aren't important to you, then nevermind.