TBR acid/base problem

[cornpops21]

If the pH of a 0.10 M weak solution is 2.70, what is the concentration of the conjugate base?

A. 0.10 M
B. 2.0 x 10^-2 M
C. 2.0 x 10^-3 M
D. 2.0 x 10^-4 M

I get answer B, but they say the answer is C. Where am I going wrong??

Here's my approach:

Since it's a weak acid solution, it does not dissociate completely, and the [H+] is equal to the [conjugate base].

Given that the pH is 2.7, I figured the Ka would be around 4 x 10^-3.

Setting up the expression, Ka = [H+][A-] / [HA] = 4 x 10^-3.

Therefore, [x][x] / 0.1 = 4 x 10^-3.

x^2 = 4 x 10^-4

x = 2 x 10^-2 = [A-]

There must be something I'm overlooking. Any help?

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[RPedigo]

...and the [H+] is equal to the [conjugate base].

Exactly. So what value of [H+] gives you pH 2.70? It's around 2 x 10^-3. You don't need to guess Ka values. All you need to do is see what [H+] (and therefore [A-]) would give you pH 2.70.

1.0 x 10^0 = pH 0
1.0 x 10^-1 = pH 1
1.0 x 10^-2 = pH 2
1.0 x 10^-3 = pH 3

So your answer must be somewhere between 1.0 x 10^-2 and 1.0 x 10^-3. Try to memorize standard log values so you can make a better educated guess where it falls between the two. Hope that makes sense

 

[Raigon]

Did you memorize the Henderson-Hasselbalch equation? I think it's easier to get the concentration of the conjugate bases from it -

Remember pH=pKa + log[A-]/[HA]

Just plug in the values and I think you can solve the problem. And like the guy above me said - memorizing basic log values help. But I think they'll give you those values on the MCATs.

And let me check to see if my answer's consistent with yours or the your answer book's.

 

[RPedigo]

Did you memorize the Henderson-Hasselbalch equation? I think it's easier to get the concentration of the conjugate bases from it -

Remember pH=pKa + log[A-]/[HA]

Just plug in the values and I think you can solve the problem. And like the guy above me said - memorizing basic log values help. But I think they'll give you those values on the MCATs.

And let me check to see if my answer's consistent with yours or the your answer book's.

The main thing here though is we are not given a Ka value, he assumed one. You should avoid doing that unless absolutely necessary, and it isn't in this particular case.

 

[Raigon]

My bad - don't use my equation above. I misunderstood your question.

you know that pH = -log[H+], right? So, your [H+] concentration is [H+]^(-ph)

= 10^(-2.7) which is equal to 10^(0.3) x 10 ^ (-3)

10^0.3, if you use log, you would figure out it's around 2.

And since [H+] equals its conjugate base concentration, it's the answer 2 x 10 ^ (-3).

Yeah, memorize basic log values. 10^0.3 = 2. You could solve for it using log.

Thanks RPedigo. =P Bleh - I'd flunk chemistry if I got something THIS basic wrong.

 

[cornpops21]

Exactly... You don't need to guess Ka values. All you need to do is see what [H+] (and therefore [A-]) would give you pH 2.70.

Haha wow. I feel like a *******. There was no need for any of my calculations.

That makes a lot of sense. Thanks!