[antoinetta]

An aqueous solution of a monoprotic acid HA (Ka = 3.68×10-1) has a pH of 3.76.
What is the molarity of the solution? (i.e. [HA]+[A-].)

So, here's what I did so far ... I found what the [H+] from the pH.  10^-3.76 = 1.737*10^-4 M

Then from the ICE table calculations:
x^2/ [HA-x] = Ka = 3.68×10-1
x= 1.737*10^-4
[HA] = 1.7308*10^-4
[A-] = x = 1.737*10^-4

So, the molarity = 1.7308*10^-4 + 1.737*10^-4 = 3.46*10^-4

Is that correct?

[Borek]

No. Check your math or something. No idea where you get wrong, but it can't be OK.

1. Ka = 0.368, this is relatively strong acid. pH is close to 4, so you can be sure acid is diluted ans almost 100% dissociated, thus its concentration must be close to 10^-3.76 which is 1.7*10^-4.

2. From your calculations [HA] is almost identical with [A^-], which means acid is dissociated 50%. That is possible if (think about Henderson-Hasselbalch equation) pKa is 3.76 - but we know pKa is about 0.4, so at this pH [HA] must differ from [A^-] by several orders of magnitude.

[AWK]

Something is wrong.

(i.e. [HA]+[A-].)

. This is an equlibrium concentration of HA.

cHA is a total concentration of an acid and should be 0.00017308 Do not use square brackets They put a misapprehension in your calculations. Your acid dissociates in 99.95 % and [H⁺] is a good approximation of the concentration of HA.

[antoinetta]

So lets try again, [H+] = Ka [HA]/[A-]

From the pH, [H+] 10^-3.76 = 1.737*10^-4 M
Ka is given = 3.68×10-1

Since it is a strong acid, then we can estimate [HA] to be 1.737*10^-4 M (is that right?)

[Borek]

Not exactly. [HA] is an equilibrium concentration of undissociated acid, you are asked about analytical concentration of the acid - so you are interested in [HA] + [A^-]. As the acid is practically 100% dissociated [A^-] = [H⁺] = 10^-3.76. If the acid is 100% dissociated its analyctical concentration Ca = [A^-] = 10^-3.76.

You may try to calculate [HA] now (using known Ka, [H⁺] and [A^-]) to check if 100% dissociation assumption holds.

[antoinetta]

ok got it I found [HA] to be 8.199*10^-8 M.

So [HA] +[A-] = 1.73*10^-4.

-----

On a side note:  Can you calculate the concentration of a strong reactant (like ammonia), reacting in a weak solution (like acetic acid) from its density? 

[english]

You do realize you are calculating the analytical molarity here?

[antoinetta]

You do realize you are calculating the analytical molarity here?

Yes.  I have already figured out the original question that I set out to ask ... it was just a side note.  It was something that I was wondering about which has nothing to do with analytical molarity.  I just want to know how to relate density with the K in relation to a strong reactant.  Sorry, if that was confusing...

[Borek]

On a side note:  Can you calculate the concentration of a strong reactant (like ammonia), reacting in a weak solution (like acetic acid) from its density?

You mean from the solution density? No, you have to look it up in the density tables.

See ammonia density table.

[antoinetta]

So, say you have this reaction:  2NH_3 <=>  NH_4 + + NH_2 -        K = 1×10-33 at -50°C

Can you still find [NH2-] in a 3.16×10-3 M solution of acetic acid in liquid ammonia at -50°C?  NH3 is very strong, but K is very small, how do I find the concentration of NH3 is ammonia?

[AWK]

Use approximate concentration of ammonia as 40.6M, and assume all acetic acid forms a salt, hence you have a problem with a common ion effect.

[Borek]

Note1: if you dissolve acetic acid in liquid ammonia you need to know its dissociation constant for calculations - and it will be completely different value than 10^-4.75 you know from water solutions.

Note2: ion products for solvents are often given with assumption that undissociated form concentration is constant, thus you don't have to know ammonia concentration. See water ion product discussion.