If I mix #25# #ml# of #0.1# #N# #HCl# with #1000# #mg# of #Mg(OH)_2# in #25# #ml# #H_2O#, what will the resulting pH be?
I really don’t understand when I can or cannot use Henderson Hasselbalch equation. If I use ICE, I realize that the base will be in excess, therefore neutralizing the HCl, but then what happens to the excess base? Which Kb will allow me to solve the pOH?
I really don’t understand when I can or cannot use Henderson Hasselbalch equation. If I use ICE, I realize that the base will be in excess, therefore neutralizing the HCl, but then what happens to the excess base? Which Kb will allow me to solve the pOH?
3 Answers
You gots strong base and strong acid....there is no question of using the buffer equation....we do use the equation
Explanation:
...we simply assess the stoichiometric equivalence according to ....
And because of this stoichiometric NON-EQUIVALENCE upon the mixture of the reagents...there are
And so
Now
But
The pH of the solution will be around
See below for the detailed solution.
Explanation:
First we need a balanced chemical equation for this neutralisation reaction:
#2HCl+Mg(OH)_2->MgCl_2+2H_2O#
Next we need to know how many moles of each reagent we have. For a monoprotic strong acid,
For
#C=n/V#
Therefore,
For
#Mg(OH)_2(s) -> Mg^(2+)(aq) + 2OH^(-)(aq)#
Its solubility is given as
#K_(sp) = 5.61 xx 10^(-12) = [Mg^(2+)][OH^(-)]^2#
#= s cdot (2s)^2 = 4s^3#
Thus,
#[OH^(-)] = 2s = 2 cdot (K_(sp)/4)^(1//3) = 2.24 xx 10^(-3)# #M# is the maximum molar solubility of
#OH^(-)# possible.
Half of this gives the molar solubility of
#1.12 xx 10^(-3) mol//L cdot 0.050# #L = 5.60 xx 10^(-6)# mols
at
#5.60 xx 10^(-6) mols xx (58.3 g)/(1 mol) = 3.26 xx 10^(-4) g#
#= 0.326# #mg# out of#1000# !!
We need
That will push the equilibrium by Le Chatelier's principle to dissociate more
The
The
That means there will be
#0.017-0.00125~~0.016# mols of#Mg(OH)_2# that is meant to dissolve. But again, it won't all dissociate.
This is in about
But implicitly from what we mentioned, only
So, the concentration will be:
#C=n/V=(1.12 xx 10^(-5))/(0.050)=2.24 xx 10^(-4)# #M#
The
#pH+pOH=14# ,
so
#pH=14-pOH#
#=14-3.65#
#~~ 10.4#
The pH will be 10.35.
Explanation:
You can use the Henderson-Hasselbalch equation only for a solution of a weak acid and its conjugate base or a solution of a weak base and its conjugate acid.
Here you have a strong acid and a strong base, so you can't use the Henderson-Hasselbalch equation.
We can use an ICE table to set up the calculations.
So, we will have a suspension of solid