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### Topic: another titration problem  (Read 10827 times)

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#### escapeartistq

• Guest ##### another titration problem
« on: May 14, 2005, 02:49:29 AM »
I am trying to solve the following problem:

0.20 g of magnesium were dissolved in 25 cm3 of HCl 1.0 M.
What is the needed volume of NAHO 0.5 M to neutralize the excess acid?

When I try to write the neutralization equation I am stuck with the Mg which keeps appearing in the products...

These are the equations I could get:

Mg reaction with HCl:

Mg2+(aq) + 2HCl(aq) -> MgCl2(aq) + 2H+(aq)

Mg reaction with H2O:

Mg2+(aq) + 2H2O(l) <=> Mg(HO)2(s) + 2H+(aq)

Mg(HO)2(s) dissolution:

Mg(HO)2(s) <=> Mg2+(aq) + 2HO-(aq)

summing the three equations I got:

Mg2+(aq) + 2HCl(aq) + 2H2O(l) -> MgCl2(aq) + 4H+(aq) + 2HO-(aq)

which I think represents the excess acid produced by reacting Mg with HCl

Now when I add the NaHO dissociation equation

2NaHO (aq) -> 2Na+(aq) + 2HO-(aq)

to the previous one it becomes a mess that I cannot simplify to obtain the usual HO-(aq) + H+(aq) -> H2O(l) water formation equation that represents the neutralization.
Maybe I am overdoing it and doing something wrong. Please *delete me*

In solving these problems I usually start by trying to write the neutralization equation so I can see clearly the stoichiometry of the equation and calculate the quantities of acid or base necessary for the neutralization.

« Last Edit: May 14, 2005, 02:54:24 AM by escapeartistq »

#### Borek ##### Re:another titration problem
« Reply #1 on: May 14, 2005, 04:07:47 AM »
Maybe I am overdoing it

Yes you do.

First thing you have done wrong is the first reaction - check the equation.

As you have to titrate excess HCl there will be no one reaction describing whole process. First will be the equation of Mg dissolving - and nothing else. Exactly what you have tried to achieve with your first reaction, just without error. Second will be the reaction of neutralization of excess HCl - and Mg2+ will be only spectators in this reaction. Forget about Mg2+ reacting with water and then dissolving (or rather try to add the equations to check what cancels out).

Ever heard about KISS rule? « Last Edit: May 14, 2005, 04:11:04 AM by Borek »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Garneck

• Guest ##### Re:another titration problem
« Reply #2 on: May 14, 2005, 09:11:56 AM »
Yeah, that HO- is hurting my eyes.. #### escapeartistq

• Guest ##### Re:another titration problem
« Reply #3 on: May 14, 2005, 12:35:52 PM »
Quote
First thing you have done wrong is the first reaction - check the equation.

Sorry I don't understand what's wrong in this equation.
How do you write the equation of Mg dissolving properly?

#### escapeartistq

• Guest ##### Re:another titration problem
« Reply #4 on: May 14, 2005, 02:02:33 PM »
Googling for the Mg reaction with HCL I found this equation
Mg(s) + 2HCl(aq) -> Mg2+(aq) + H2(g) + 2Cl-(aq)

but I am still lost. Because I don't understand how reacting Mg(s) with HCL(aq) gives excess acid.

#### Borek ##### Re:another titration problem
« Reply #5 on: May 14, 2005, 02:13:53 PM »
You use initially more acid than is needed. After Mg is dissolved some of the acid is still in the solution. Titration is for checking how much acid was left.

Similar titration can be used for determination of Mg mass -  inital (known) amount of acid minus amount of acid left (titrated) is the amount of acid that reacted with Mg - this way you can determine what was amount of Mg.
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#### Garneck

• Guest ##### Re:another titration problem
« Reply #6 on: May 14, 2005, 02:39:59 PM »
Googling for the Mg reaction with HCL I found this equation
Mg(s) + 2HCl(aq) -> Mg2+(aq) + H2(g) + 2Cl-(aq)

but I am still lost. Because I don't understand how reacting Mg(s) with HCL(aq) gives excess acid.

Where do you have excess acid? It's a simple redox reaction. Magnesium oxidises to Mg2+, H+ gets reduced to hydrogen:

Mg(s) + 2H+(aq) + 2Cl-(aq) -> Mg2+(aq) + H2(g) + 2Cl-(aq)

So if you remove the chloride ions from the both sides of the reaction, you get:
Mg(s) + 2H+(aq)  -> Mg2+(aq) + H2(g)

#### escapeartistq

• Guest ##### Re:another titration problem
« Reply #7 on: May 14, 2005, 04:02:57 PM »
thank guys!  Now I understand how to solve the problem.
« Last Edit: May 14, 2005, 04:06:14 PM by escapeartistq »

#### GCT

• Guest ##### Re:another titration problem
« Reply #8 on: May 14, 2005, 04:28:25 PM »
thank guys!  Now I understand how to solve the problem.

Mg2+ will act as an acid...try figuring out how.

#### GCT

• Guest ##### Re:another titration problem
« Reply #9 on: May 14, 2005, 04:45:19 PM »
thank guys!  Now I understand how to solve the problem.

1)Mg2+(aq) + 2HCl(aq) -> MgCl2(aq) + 2H+(aq)

this reaction is incorrect, the net ionic equation cancels out completely, no net reaction occurs

2)g2+(aq) + 2H2O(l) <=> Mg(HO)2(s) + 2H+(aq)

This reaction is what you should be using.

3)Mg(HO)2(s) <=> Mg2+(aq) + 2HO-(aq)

This one is superflous since we it is indicated in the second equation that equilibrium in the formation of the solid has been achieved.

Thus the net neutralization reaction is

2H30+ + 20H- --->4H20(l)
=H30+ + 0H- ---->2H20(l)

the hydronium is from equation 2.

all of this, assuming that your equations were balanced properly
« Last Edit: May 14, 2005, 04:46:17 PM by GCT »

#### escapeartistq

• Guest ##### Re:another titration problem
« Reply #10 on: May 14, 2005, 06:10:39 PM »
Quote
this reaction is incorrect, the net ionic equation cancels out completely, no net reaction occurs

So you say that first I should use

1) Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

and then

2) Mg2+(aq) + 2H2O(l) <=> Mg(OH)2(s) + 2H+(aq)

in which Mg2+ is a Lewis Acid, according to the information here http://www.meta-synthesis.com/webbook/12_lab/lab.html

Is this ok?

BTW, I believe the recommended nomenclature for the hydroxide ion is HO-, and not OH- since 1957, but I am a newbie in these matters

#### Borek ##### Re:another titration problem
« Reply #11 on: May 14, 2005, 06:26:00 PM »
So you say that first I should use

1) Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

GCT is wrong, there is no need for the second reaction. There is too much HCl in solution used for dissolving Mg, so after all Mg is dissolved some of the acid is left and you must titrate this excess with NaOH.

Quote
BTW, I believe the recommended nomenclature for the hydroxide ion is HO-, and not OH- since 1957, but I am a newbie in these matters

It is OH-.
« Last Edit: May 14, 2005, 06:29:16 PM by Borek »
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#### GCT

• Guest ##### Re:another titration problem
« Reply #12 on: May 15, 2005, 01:55:31 PM »
So you say that first I should use

1) Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)

and then

2) Mg2+(aq) + 2H2O(l) <=> Mg(OH)2(s) + 2H+(aq)

in which Mg2+ is a Lewis Acid, according to the information here http://www.meta-synthesis.com/webbook/12_lab/lab.html

Is this ok?

BTW, I believe the recommended nomenclature for the hydroxide ion is HO-, and not OH- since 1957, but I am a newbie in these matters

nice web page by the way...although it takes forever to load due to all of the pics.

Yes, I believe that the reactions you had in the original posts are valid.  And in this case, the product of magnesium solid and aqueous hydrochloric acid is aqueous magnesium chloride.  By aqueous it means that magnesium is a spectator ion, dissolved in water...in the form of Mg2+.

As you can see, this may undergo further reactions with water, contributing to an acidic pH.  From this you can deduce the stoichiometric ratios to your advantage, depending on which ratio you need.

However, the answer depends on the original question, I wasn't absolultely sure if Mg2+ would be reactive enough as a lewis acid, however, according to the webpage you provided it is.