[THC]

Hi

I'm having some difficulties with this one:

60,0 mL 1,50 M NH3 (aq) is added to 15,0 mL 0,100 M MgCl2 (aq). Show that Mg(OH)2 precipitates (is that the right word?).


My answer:
The reactions:
NH3 + H2O <-> NH4+ + OH-
Mg2+ + 2 OH- -> Mg(OH)2

I calculated [OH-] = 10^[-(14-½{4,75 - log 1,2})] = 2,16*10^-12 M.
Besides that, [Mg2+] = 0,02 M, so Ks = 0,02 M*(2,16*10^-12 M)^2 = 9,37*10^-26.
The theoretical Ks is 1,2*10^-11, but according to my calculations, there's no precipitate. What am I doing wrong?
Thanks.

[Borek]

I calculated [OH-] = 10^[-(14-½{4,75 - log 1,2})] = 2,16*10^-12 M

Value seems to fit H⁺ concentration in this solution, not OH^-.

Have you checked if the approximate formula can be used?

[THC]

I calculated [OH-] = 10^[-(14-½{4,75 - log 1,2})] = 2,16*10^-12 M

Value seems to fit H⁺ concentration in this solution, not OH^-.

Have you checked if the approximate formula can be used?

!!! Yes, you are right. It is [OH-] = 10^[-½{4,75 - log 1,2}].
That was stupid Thanks again.

What is the approximate formula?

[AWK]

Hi

I'm having some difficulties with this one:

60,0 mL 1,50 M NH3 (aq) is added to 15,0 mL 0,100 M MgCl2 (aq). Show that Mg(OH)2 precipitates (is that the right word?).


My answer:
The reactions:
NH3 + H2O <-> NH4+ + OH-
Mg2+ + 2 OH- -> Mg(OH)2

I calculated [OH-] = 10^[-(14-½{4,75 - log 1,2})] = 2,16*10^-12 M.
Besides that, [Mg2+] = 0,02 M, so Ks = 0,02 M*(2,16*10^-12 M)^2 = 9,37*10^-26.
The theoretical Ks is 1,2*10^-11, but according to my calculations, there's no precipitate. What am I doing wrong?
Thanks.

Why you neglected formation of ammonium buffer with concentration 1,16 M of NH₃ and 0.04 M NH₄Cl

[Borek]

Why you neglected formation of ammonium buffer with concentration 1,16 M of NH₃ and 0.04 M NH₄Cl

Where do you see NH₄Cl?

[Borek]

What is the approximate formula?

The one you have used - see equation 8.13.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[THC]

What is the approximate formula?

The one you have used - see equation 8.13.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Well, pKb is 4.76, so it should be alright to use that formula, right?

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Well, pKb is 4.76, so it should be alright to use that formula, right?

It is OK to use this formula in this case, but it is not only pKb dependent. Reread the page, take a look at the plot.

[THC]

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Well, pKb is 4.76, so it should be alright to use that formula, right?

It is OK to use this formula in this case, but it is not only pKb dependent. Reread the page, take a look at the plot.

I did take c_b into account, if that's what you're suggesting. I used Kb = c_b*a^2/(1-a) where a = [OH-]/cb. Then I calculated for c_b = 1.2 M and a = 0.05 (from the link). This yields pKb = 2.5 (AFAIR), so it should be quite safe to use the approximate equation with pKb = 4.76.

Other than that, we've been told to use pH = ½(pKb - log c_b) for bases (or the equivalent with acids) if pKb > 4 :-)

[Borek]

So you were told wrong, try to calculate pH of 10^-5 M ammonia solution using approximation, and remember that the real value is 8.85. This approximation can be used as long as dissociation fraction is below 5%.

See

http://www.chembuddy.com/?left=BATE&right=pH-cheat-sheet

for a sure way of calculating pH with the best approximations possible.