Two solutions of different concentrations of acid are mixed creating 40 mL of a solution that is 32% acid. One-quarter o...

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"Two solutions of different concentrations of acid are mixed creating 40 ml of a solution that is 32% acid. one-quarter of the solution is made up of a 20% acid solution. the remaining three-quarters is made up of a solution of unknown concentration, c. equation can be used to determine c, the unknown concentration?

a) 30c + 10(0.2) = 40(0.32)

b) (c) + (0.2) = 40(0.32)

c) (c)( (0.2)) = 40(0.32)

d) 30(c)(10(0.2)) = 40(0.32)"

With this, we can now solve the question.

We have a solution that was made mixing2 different solutions of acid. In this case, we can assume the following:

Solution 1 = 40 mL 32% acid

Solution 2 = 1/4 solution 1, 20% acid

Solution 3 = 3/4 solution 1, c% acid.

Thus we can say:

Solution 1 = Solution 2 + Solution 3     (1)

Now, 1/4 of the volume of solution 1, means that we use 10 mL of a 20%acid, and the remaining 30 mL is the unknown acid.

With this, we can discart option b) and c) because these option are not considering the volume of solution 3 used. Option d is not doing any sum, only multiplications, so, this is not the option to use. We can easily say it's option a, but let's prove it.

In general terms, concentrations and volume are related, and if we have two solutions, the information of these two solutions must be conserved in the resulting solution. In other words:

M₁V₁ = M₂V₂

According to this, the concentration of the unknown acid would be:

30c + 0.2*10 = 40(0.32)

30c + 2 = 12.8

30c = 10.8

c = 0.36

Which means that the concentration of the acid is 36%

Hope this helps

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