A chemist has on hand a
supply of a 50% methyl alchohol solution and a 20% methyl alcohol
solution. How much of each kind should he mix to produce 300 grams of a
30% methyl alcohol solution?
The micro-world
of Solutions
Solution problems are
"chemistry-like" problems. They are frequently stated using the
terms solution and concentration (or strength),
and may refer toconcentration as a percent (%)
solution (of the principal ingredient).
To make these concepts
concrete, consider the following simple procedure: we mix 8 oz. of water with 2
oz. of salt to obtain a solution of salt in water (commonly called
"saltwater" or "brine"):
In this picture (as in
all to follow), we show the constituents of the mixture as being
"separated" so we can see their numerical relationshops. Of
course, in a real brine solution, the salt and water would be completely mixed
and indistinguishable one from the other.
The salt is referred to
as the solute, the water is the solvent, and the
resulting mixture as the solution. We need a way of quantifying
the concentration of salt (that is, a number that expresses
how salty the salt water really is). "Concentration" is defined
to be
the ratio of the amount of solute (salt),
to the total amount of solution (water + salt)
In this case, the total
amount of solution is 8 + 2 = 10 oz. Since the amount of salt is 2 oz.,
the concentration of salt in this solution will be 2/10 = .20 (or 20%).
Note that concentration
is usually expressed as a percent (obtained from the decimal concentration
by multiplying by 100). In general, in the statement of
problems and their answers, concentrations will be expressed using %. But in
any calculations involving concentrations, we will always use
the decimal form of the concentration.
In the next example, we
are presented with 100 grams of a 20% solution of salt in water:
It is reasonable to ask
the question "How much salt is there in this solution?".
Mathematically, we are asking: "What is 20% of 100?". With the
reminder that "of" usually is translated into the mathematical operation
of "times" (multiplication), we then refine the question to
"What is 20% times 100?". Finally, by converting the 20%
to decimal form, we can answer the original question:
amount of salt = (.20)(100) = 20 grams
We can compute the
amount of salt (solute) in the solution by multiplying the concentration (converted
to decimal form by dividing by 100) by the total amount of
solution.
Note that the solute in
any solution problem can be identified as that component of the solution
referred to using the % sign; for example, if a problem refers to a 7% solution
of pure hydrochloric acid in water, the hydrochloric acid is the solute, and
the water is the solvent.
Concentration will
always be a number on the range from 0% - 100% (0.0 - 1.0 in decimal form).
A concentration of 0%
(0.0) represents the absence of any solute. We can, if we wish, refer
to pure water as a 0% concentration of salt in water!
On the other hand, a
100% solution represents pure solute (no solvent). A 100% solution of
ethyl alcohol is pure (undiluted) ethyl alcohol.
Our calculational
formulas still hold in these extreme cases: if we have 13 oz. of a 100%
solution of anti-freeze, the amount of anti-freeze is given by
(1.0)(13) = 13 oz. of anti-freeze (naturally!). On the other hand, 13 oz.
of a 0% solution of anti-freeze contains (0.0)(13) = 0 oz. of anti-freeze
(naturally!).
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Word
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Definition
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solution
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a mixture of two chemicals or
substances
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solute
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that component of a solution that
is referred to by its concentration (strength), usually as a %
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solvent
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the non-solute component of the
solution, sometimes water
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concentration
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a measure of the strength of the
solute in the solution, given by the formula:
(concentration) =(amount of solute)
-------------------
(total amount of solution)
The resulting number is usually
expressed as a % by multiplying by 100.
The mathematical model
for solutions
According to the above
discussion, we measure concentration by the following formula:
(concentration) =
-------------------
(total amount of solution)
By rearranging this equation,
we have
(amount of solute) = (concentration)(total amount of solution)
It was this form of the
math model that was used in the opening discussion to compute the amount of
salt in 100 grams of a 20% sulution of salt in water:
amount of salt = (.20)(100) = 20 grams
This is the math model
that will be used extensively in what follows. Concentration as
used in this formula (and in any equation) will be the decimal form of
concentration (a number on the range 0.0 - 1.0).
So far, we have
discussed situations involving only one solution at a time. Most solution
word problems create a little more interest by referring to two (or more)
solutions of different concentrations that are combined to
form a new solution of yet another concentration.
For example, consider
the following scenario:
10 oz. of a 20% solution
of salt in water (solution A) is combined with 12 oz of a 50% solution of salt
in water (solution B) to create a combined solution (solution C).
It is not too hard to
figure out the amount of salt in solution C:
(amount of salt in solution C) = (amount of salt in solution A) +
(amount of salt in solution B) =
(.20)(10) + (.50)(12) = 2 + 6 = 8 oz. of salt
In fact, we can also
compute the concentration of salt in solution C. Since
the total amount of solution C is 10 + 12 = 22 oz., of which 8
oz. is salt (we just computed that), we have:
(concentration of salt in solution C) = 8/22 = .36
(36%)
Summing up, we can state
that the combined solution (solution C) consists of 22 oz. of a 36% solution of
salt in water.
Consider our opening
problem:
A chemist has on hand a
supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How
much of each kind should he mix to produce 300 grams of 30% methyl alcohol
soluton?
There is a fundamental
property possesed by this problem situation (and all others like it). We
have already implicitly used this property in the above discussion
"Combining Solutions". Note well: the first component of the solution
has a certain amount of alcohol, the second component has a certain amount of
alcohol, and we are combining these two solutions. It stands to reason
that, since this process neither creates or destroys alcohol,
the amount of alcohol in the combined solution
will be
the sum of the amounts of alcohol in the constituent solutions
This observation is
basic to the solution of all solution problems, and forms the
basis for creating the Solution Problem Rough Equation. We call it
the
Solute Equation
The corresponding rough
equation for our opening problem will be:
(amt. alcohol in the 50% sol.) + (amt. alcohol in the 20% sol.) =
(amt. alcohol in the 30% sol.)
In the above exercises,
you were able to use the math model to compute concentration (or, in its other
form, to compute the amount of solute) in situations where only numerical
quantities were involved. In the solution of algebra problems, the various
elements of the problem are not always expressible numerically,
because of the presence of an "unknown". Consider again our
opening problem:
A chemist has on hand a
supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How
much of each kind should he mix to produce 300 grams of 30% methyl alcohol
soluton?
The final solution of
this problem will start with a statement that identifies one of the unknowns of
the problem, for example:
let x = number of oz. of 50% alcohol
solution to be used
Accordingly, in the
course of solving the problem, x can be used, wherever
it is needed, to represent the amount of 50% alcohol to be used. But how
much pure alcohol is in this x ounces
of 50% alcohol? Of course, if we knew that x =
100, we could readily compute the amount of pure alcohol: (.50)(100) = 50 oz.
of alcohol. But we don't know what x is
(after all, it is the unknown!). The best we can do is multiply .50
times the x and express the result as an algebraic
expression, not a number. So what we get is
(.50)(x) = .50x (oz.
of pure alcohol in x ounces of 50% alcohol solution)
This may not seem very
informative, but it is:
(1) correct (by virtue of the math model)
(2) the best we can do for now (3) very handy in finding out later exactly what x is!
Unknowns can be combined
in other ways to create "correct" algebraic expressions to be used in
our "calculations".
Suppose we mix 10 oz. of
solution A with a certain amount of solution B to obtain 30 oz. of the combined
solution. How much solution B was used? I think we would all agree that the
answer is 20 oz. How did you get that answer? You probably did it very
quickly, almost without thinking about it, as follows:
(amt. of solution B) = (combined amt.) - (amt. of solution A) = 30
- 10 = 20 oz.
On the other hand,
suppose we don't know the actual amount of solution A -- only that it is an
unknown which we have called x. How do we now answer the
above question?
By the very same process, except that now we
cannot actually do the subtraction, but only indicate it symbolically.
So the answer is:
30
- x (oz.)
This is correct,
and the best we can do.
Returning again to our
opening problem:
A chemist has on hand a
supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How
much of each kind should he mix to produce 300 grams of 30% methyl alcohol
soluton?
As before, we start
with:
let x = amount of 50%
alcohol to be used
Now we can write an
expression for the amount of 20% alcohol to be used:
300 - x
(grams)
Solving solution
problems
With the above
preparation, we are ready to attack our opening problem, and see it through to
completion:
A chemist has on hand a
supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How
much of each kind should he mix to produce 300 grams of 30% methyl alcohol soluton?
Step 1: Identify what it is we are looking for. According to the problem statement, we need to identify two things:
(1) the amount of 50%
alcohol, and
(2) the amount of 20% alcohol to be combined to form the 30% solution.
We will be using a method
of solution that involves only one unknown. Since one
unknown cannot stand for two things at once, we have to choose (the choice is
arbitrary). I made the choice:
#1: let x = the amount
of 50% alcohol to be used
Step 2: Draw a picture for problems involving lengths or distances. This problem does not, so:
#2: N/A
Step 3: (1) Write the rough equation, based on the Solute Equation, as stated above:
#3:
rough equation: (amount of alcohol in 50% solution) + (amount of alcohol in 20% solution)
= (amount of alcohol in 30% solution)
(2) Refine the
rough equation into a finished equation. We do so by making
repeated use of the math model:
(amount
of solute) = (concentration)(total amount of solution)
This will turn our rough
equation into the following:
(conc. of 50%
solution)(amt. of 50% solution) + (conc. of 20% solution)(amt. of 20% solution)
=
(conc. of 30% solution)(amt. of 30% solution)
We are now ready to
refine further, by making use of things we know:
(1) The various
concentrations are easy: .50, .20, and .30 respectively.
(2) The amount of 50% solution is also easy: it is simply x (see the "let statement"!). (3) But what about the amount of 20% solution? Since the total amount is to be 300 ozs., that leaves 300 - x for the amount of 20% solution.
Assembling all this
information, we arrive at the
finished equation:
.50x + .20(300 - x) = .30(300)
Step 4: Solve and check the finished equation:
#4:
.50x + (.20)(300) - .20x = 90
.30x = 90 - 60 .30x = 30 x = 100 check: .50(100) + (.20)(300 - 100) ?= .30(300) 50 + 40 ?= 90 (it checks) Step 5: State the answer.
Since the question was
"how much of each solution must we mix", we must provide two
"answers" . The answer to the amount of 50% solution is given by the
solution to our equation: x = 100 oz.. What about
the amount of 20% solution? Since there is to be 300 oz. altogether, this
leaves 200 oz. for the 20% solution. Our answer looks like this:
#5:
100 oz. of 50% methyl alcohol, 200 oz. of 30% methyl alcohol.
Note that we can check
our answer by doing the following calculations:
amt. alcohol in 50% solution: 50% of 100 oz. =
50 oz of alcohol
amt. alcohol in 20% solution: 20% of 200 oz. = 40 oz. of alcohol for a total of 90 oz of alcohol in the resulting mixture.
What will be the
concentration of alcohol in the resulting mixture? It will be
90/300 = .30 (30%)
which satisfies the
conditions of the problem, that we are to create 300 oz. of a 30% solution.
We repeat the whole
solution as it should be written out by the solver:
A chemist has on hand a
supply of a 50% methyl alchohol solution and 20% methyl alcohol solution. How
much of each kind should he mix to produce 300 grams of 30% methyl alcohol
soluton?
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