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?

3 Answers
Jan 2, 2018

You gots strong base and strong acid....there is no question of using the buffer equation....we do use the equation #14=pH+pOH#

Explanation:

...we simply assess the stoichiometric equivalence according to ....

#Mg(OH)_2(s) + 2HCl(aq) rarr MgCl_2(aq) + 2H_2O(l)#

#"Moles of magnesium hydroxide"=(1000*mgxx10^-3*g*mg^-1)/(58.32*g*mol^-1)=0.0172*mol#

#"Moles of hydrochloric acid"=25*mLxx10^-3*L*mL^-1=0.0172*mol#

And because of this stoichiometric NON-EQUIVALENCE upon the mixture of the reagents...there are #0.0172*mol# with respect to #Mg(OH)Cl(aq)# dissolved in a #50*mL# volume of solution...

And so #[HO^-]=(0.0172*mol)/(0.050*L)=0.344*mol*L^-1#

Now #pOH=-log_10[HO^-]#...here #pOH=0.46#

But #pH=14-pOH=14-0.46=13.54#

The pH of the solution will be around #10.4#.

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, #0.1# #N# is the same as #0.1# #M#.

For #HCl#:

#C=n/V#

Therefore, #n=CV=0.1xx0.025=0.0025# mols of #HCl#

For #Mg(OH)_2#, it is NOT a strong base, and has a #K_(sp)# of about #5.61 xx 10^(-12)#:

#Mg(OH)_2(s) -> Mg^(2+)(aq) + 2OH^(-)(aq)#

Its solubility is given as #s# in the mass action expression:

#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 #Mg(OH)_2(aq)#. #Mg(OH)_2# has a molar mass of approx. #58.3# #g cdot mol^-1#, so in #50# #mL# of water, we manage to dissolve

#1.12 xx 10^(-3) mol//L cdot 0.050# #L = 5.60 xx 10^(-6)# mols

at #25^@ C#, or only...

#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 #2# mols of #HCl# for each mol of #Mg(OH)_2#, but clearly the #0.326# #mg# of dissolved #Mg(OH)_2# is completely neutralized.

That will push the equilibrium by Le Chatelier's principle to dissociate more #OH^(-)#, until all the #OH^(-)# is finally dissolved and dissociated.

The #1000# #mg# from similar calculations is equal to #0.017# mols, so it will come to the point where there will be leftover #Mg(OH)_2#, and we will be looking for a pH between 7 and 14 at #25^@ C#.

The #0.0025# mol of #HCl# will react with half as much, #0.00125# #mol#, of #Mg(OH)_2# (according to the balanced reaction).

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 #50# #mL# of solution, assuming the volumes are additive.

But implicitly from what we mentioned, only #1.12 xx 10^(-5)# mols of #OH^(-)# (twice compared to the #Mg^(2+)#) will dissociate.

So, the concentration will be:

#C=n/V=(1.12 xx 10^(-5))/(0.050)=2.24 xx 10^(-4)# #M#

The #pOH# is given by #-log_(10)# of this concentration, which is #3.65#.

#pH+pOH=14#,

so

#pH=14-pOH#

#=14-3.65#

#~~ 10.4#

Jan 3, 2018

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.

#M_text(r):color(white)(mmmmmmmll)58.32#
#color(white)(mmmmm)"2HCl" + "Mg(OH)"_2 → "MgCl"_2 + "2H"_2"O"#
#"I/eq": color(white)(mll)0.0025 color(white)(ml)0.034 29"#
#"C/eq": color(white)(m)"-"0.0025color(white)(m)"-"0.0025color#
#"E/eq": color(white)(mmm)0color(white)(mml)0.0318#

#"Eq. of HCl" = 0.025 color(red)(cancel(color(black)("L HCl"))) × "0.1 eq HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.0025 eq HCl"#

#"Eq. of Mg(OH)"_2 = 1.000 color(red)(cancel(color(black)("g Mg(OH)"_2))) × (1 color(red)(cancel(color(black)("mol Mg(OH)"_2))))/(58.32 color(red)(cancel(color(black)("g Mg(OH)"_2)))) × ("2 eq Mg(OH)"_2)/(1 color(red)(cancel(color(black)("mol Mg(OH)"_2)))) = "0.034 29 eq Mg(OH)"_2#

So, we will have a suspension of solid #"Mg(OH)"_2#.

#color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + 2"OH"^"-""(aq)"#
#"E/mol·L"^"-1": color(white)(mmmmmmmmmm)xcolor(white)(mmmmmll)2x#

#K_text(sp) = ["Mg"^"2+"]["OH"^"-"]^2 = x(2x)^2 = 4x^3 = 5.61 × 10^"-12"#

#x^3 = (5.61 × 10^"-12")/4 = 1.40 ×10^"-12"#

#x = 1.12 × 10^"-4"#

#["OH"^"-"] = 2xcolor(white)(l) "mol/L" = 2 × 1.12 × 10^"-4" color(white)(l)"mol/L" = 2.24 × 10^"-4"color(white)(l) "mol/L"#

#"pOH" = "-log"["OH"^"-"] = "-log"(2.24 × 10^"-4") = 3.65#

#"pH = 14.00 - 3.65 = 10.35"#