. Mg reacts differently with ammonium hydroxide than it does with combo. Fe* rea
ID: 697991 • Letter: #
Question
. Mg reacts differently with ammonium hydroxide than it does with combo. Fe* reacts in the same way with ammonium hydroxide as it does with combo. Combo is a mixture (What is this kind of mixture called?) which is 0.30 M ammonium hydroxide/ammonia and 3.0 M ammonium nitrate. Benchtop ammonia is 3.0 M Calculate [OH] in 3.0 M NH3. Kb(NH3) = 1.74 x 105 b. Calculate [OH'] in combo. (Note you will have to calculate pK.(NH) to use the Henderson-Hasselbalch equation.) c. Use your results from 1b, 2a and 2b to explain the difference in behavior of the Mg#2 and the Fe3 ions with the two reagents.Explanation / Answer
That "combo" is also called PH buffer
A) The reaction you want is
NH3 + H2O ===== NH4 + OH
here we have to remember the K equilibrium equation
K = [products] / [reactants] , the [ ] means the concentration of something so
Kb = [OH ] [NH4] / [NH3] = 1.74 x 10-5 we have to make an ICE chart
NH3 + H2O ===== NH4 + OH
0.3...........................0...........0
-x.............................+x.........+x
3-x........................x.............x
Kb =x2 / (3 - x) = 1.74 x 10-5
if you solve this equation you will get a value for x = 0.00721, this is the concentration of OH so
b) For the buffer solution we can use the handerson hasselbach equation like
POH = Pkb + log (NH4+ / NH3)
POH = -log(1.74x10-5) + log ( 0.3 / 3 )
POH = 4.76 - 1 = 3.76
[OH] = 10-3.76 = 0.000173 M
C) the difference comes from the the solubilities, remember that for a precipitate like Mg(OH)2 and Fe(OH)3 the ph will have an effect, from the experiment we can say that the buffer has an OH below the limit required to produce the precipitate
For Fe(OH)3 we can say that the concentration of OH is enough to make a precipitate
This can also be concluded by taking a look to the ksp for each compound, the value of Ksp for Fe(OH)3 is much lower than the one for Mg(OH)2 this means that the Fe requires less ammount of oh to precipitate so this compound will form before than Mg(OH)2
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