Shameful problem
 

Hi people,

The adding of ammonium (NH4+) to water causes the pH to drop, right? I've been searching for a way to find the pH of a solution which contains respectively 1.8E-5, 7.1E-6, and 3.6E-6 mole/L ammonia, and found that the pH's were respectively 7.001; 7.19 and 7.35... There must be something wrong with my method, since even if it would cause the pH to rise, then it would rise together with the concentration, and not rise while concentration drops. Could anyone propose a correct solution to this problem?

A very confused (and not half ashamed) student

acidity
 

you are discussing two different things - ammonia ( a base) and ammonium (an ion)

addition of an ammonium salt (of a strong acid i.e. ammonium chlopride ) will cause the pH to drop (i.e. get smaller) but addition of ammonia (a base) will cause the pH to get larger

which would you like explained?

I am talking about the addition about ammonium to water; by using the Ka value (9.25), i can theoretically calculate the concentration of hydrogen atoms by using a quadratical equation, is this possible? By using this equation i get pH values which are higher instead of lower than 7 which is very disturbing indeed. Any help would be appreciated :)

PS: not sure if ammonium can be qualified as a strong acid

Going back to your original question, if you wanted a solution of 1.8 x 10-5 M [NH3], the best way would be to find a solution of concentrated NH3 (ammonium hydroxide) and use the dilution equation to solve for how much water you must dilute the solution with (it will be a lot).
M1V1 = M2V2

Analytically, and using your first number (1.8 x 10-5 M NH3)
NH3 + H2O --> NH4+ + OH-
(1.8 x 10-5 - x) x x x

Kb = 1.8 x 10-5 = X2 / (1.8 x 10-5 - X)
Solving for X yields OH-
determine H+ using the dilute solution relationship (H+)(OH-) = 1 10-14
determine pH - value should be in the 8-9 region

The lower concentrations will be closer to 7. The addition of something like ammonium chloride (the conjugate acid of ammonia) will cause the pH to drop.

no ammonium is a weak acid. First off, I think that 9.25 is actually the pKa, you'll need to inverse log this value to obtain the ka. The rest is simple

Ka=[NH3+x][H30+ +x]/[initial ammonium-x]

=[x][x]/[initial ammonium concentration-x], solve for x, which is the acid concentration (hydronium)

how would you obtain the pH from the hydronium concentration?

Thanks a bunch for the proposed solutions, but apparently they are both wrong. Before i posted here, i used the quadratic equation proposed by GCT; but that gave me very strange pH's indeed. It seems like the concentrations are so low that autodissociation of water (H2O ---> H3O+ +OH-) should be taken into consideration. I eventually found the solution in one of my former chemistry courses, so no need to pay my problem further heed. Nevertheless i'm grateful for your quick replies. Keep it up!

If you performed the calculations correctly, it should have given you the answer. Did you use the x to find the pH, that is did you take that negative logarithm of x? Did you take the inverse log of the pKa value to obtain Ka. This is not a difficult problem at all.

I've never had to consider the autodissociation of water in calculations, unless you are dealing with a very, very, very small initial concentration of ammonium. It is ammonium right? Note that I was replying for the ammonium suggestion (in some cases you were wondering about ammonia, you know that there's a difference right?).

The "very strange pH" indicates that you've failed to solve correctly for the solution. Suggest to me an initial concentration of ammonium, and I'll tell you the pH in a jiffy. You should NOT have to take into account the autodissociation of water in most cases.

How does the jiffy affect the pH?

Re: Shameful problem
 

9.04, 8.74 and 8.49

BATE rulez 8)

Best,
Borek

Smarkotan oz gluthozmaz