Question

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?

Answer 1

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`

Answer 2

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`

Answer 3

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"`