| 5 Fe2+ (aq) | MnO41- (aq) | 8 H3O1+ (aq) | -> | 5 Fe3+ (aq) | Mn2+ (aq) | 12 H2O | |
| start | 100 mL | 100 mL | |||||
| Concentration | 0.500 M | 0.100 M | excess | ||||
| start | 0.0500 mole | 0.0100 mole | |||||
| use | 0.0500 mole | 0.0100 mole | 0.0800 mole | ||||
| produce | 0.0500 mole | 0.0100 mole | 0.120 mole | ||||
| Final | 0 mole | 0 mole | ? | 0.0500 mole | 0.0100 mole | ? | |
| Volume | 200 mL | ||||||
| Concentration | 0.250 M | 0.0500 M | ? |
| KHPh | NaOH | -> | H2O | K1+ | Ph2- | Na1+ | |
| Given (KHPh) | 0.6945 g KHPh | ||||||
| MW (KHPh) | 204.3 g mole-1 | ||||||
| Moles (KHPh) | 3.399x10-3 mole | ||||||
| Moles NaOH | 3.399x10-3 mole | ||||||
| Volume NaOH | 39.3 mL | ||||||
| Concentration NaOH | 8.65x10-2 mole liter-1 |
| Mg(OH)2 | 2 HNO3 | -> | 2 H2O | Mg(NO3)2 | |
| Given | 0.4563 g | ||||
| MW | 58.32 g mole-1 | ||||
| Given | 0.007824 mole | ||||
| HNO3 | 0.01565 mole | ||||
| Volume HNO3 | 45.63 mL | ||||
| [HNO3] | 0.3429 mole liter-1 |
A base solution is made by dissolving 4.987 g of Potassium Hydroxide in 500.0 ml of water. 36.84 ml of this base solution is used to titrate 25.00 ml of sulfuric acid. Write the total, total ionic, and net ionic equations for these reactions. What can you calculate from this information? What are your results?
Chem 145,
Included is the solution to the acid-base problem from class
today. What I am going to do here is describe how I would solve
the problem. This is not necessarily the only way, but I want to
show you some tools for solving word problems like this (which
seem to be causing a lot of you some difficulty). These problems
are more difficult to solve than problems that just ask you to
solve for a number in a certain setup. The downside to working
problems with just the numbers, is that life is not like that. I
consider learning how to solve this type of problem to be an
important part of this course. If you can solve this question,
you really do have a strong understanding of the chemistry
involved here. That is more than just an ability to complete the
algebraic manipulations.
SOLUTIONS:
A base solution is made by dissolving 4.987 g of Potassium
Hydroxide in 500.0 ml of water. 36.84 ml of this base solution is
used to titrate 25.00 ml of sulfuric acid. Write the total, total
ionic, and net ionic equations for these reactions. What can you
calculate from this information? What are your results?
In finding out what is going on here the first thing I need to
do is decide what the reaction is. The compounds given here are
Potassium Hydroxide (the formula for this is KOH, I recognize the
hydroxide or OH part is a base) and sulfuric acid (the formula
for this is H_2SO_4, which I recognize as an acid with TWO
protons). If I am not certain of the total equation, I'll start
with some pieces:
First, dissolving KOH in water. Since KOH is a strong base I
know it completely dissociates in water. That means that when I
dissolve the KOH it will break apart. Recognizing that they OH is
a stable piece, this will happen as
KOH(s) -> K1+ (aq) + OH1-
(aq)
Doing the same for the sulfuric acid, and recalling it is
DIPROTIC:
H2SO4 -> 2 H1+ (aq)
+ SO42- (aq)
Now I can combine all these ions in solution and have
available (THIS IS NOT BALANCED YET):
K1+ (aq) + OH1- (aq) +
2 H1+ (aq) + SO42- (aq)
Looking at these and playing around with combinations I see
that it is possible to make H2O (This is good since it
is an acid base reaction) and K2SO4. Since
I remember that alkali metal sulfates are soluble (and If I do
not remember, I at least know where to look it up) K22SO4
will not form a precipitate, it will stay in solution as ions, so
these are spectator Ions. Now I can write a balanced total ionic
equation:
2K1+ (aq) + 2OH1- (aq) + 2H1+ (aq) + SO42- (aq) -->
2H2O + 2K1+ (aq) + SO42-
(aq)
The net ionic equation for this is:
2 OH1-(aq) + 2 H1+ (aq)
--> 2 H2O
And the total equation for this is:
2 KOH (aq)+ H2SO4 (aq)
--> 2 H2O (l)+K2SO4
(aq)
Now That I know what is going on, I can start to do some
calculations:
The solution is made by dissolving 4.987 g of Potassium
Hydroxide in 500.0 ml of water.
Since the molecular weight of potassium hydroxide is 56.10
g/mol this corresponds to 0.08889 moles of KOH
Since this is dissolved in 500.0 ml of water the concentration of the solution is:
(0.08889 moles)/(0.5000 L) = 0.1778 M KOH
36.84 ml of this base solution is used to titrate 25.00 ml of
sulfuric acid.
In 36.84 ml of the KOH solution there will be:
(0.1778 mol/L)*(0.03684 L) = 6.549*10-3 moles
Since the reaction of KOH with H2SO4 is
in a 2:1 ratio, the amount of H2SO4 that
will react is:
6.549*10-3 mole KOH *[(1mol H2SO4)/(2
mol KOH)] =
3.275 * 10-3 moles H2SO4
Since this number of moles of H2SO4 is
present in 25.00 ml of solution the concentration is:
(3.275*10-3 moles)/(0.02500 L) = 0.1310 M
Also since the molecular weight of H2SO4
is 98.07 g/mol the number of grams of H2SO4
is calculated as:
3.275*10-3 mol * 98.07 g/mol = 0.3212 grams H2SO4
From the above information this can be reorganized into a
table, Like We have been using in class. I will leave that
exercise to you. This is to show you another way to think about
solving this type of problem. The important point is to look
carefully at the information that is provided and find a way to
sort through and organize it. If you have any questions about
this, please let me know.