The caption of solution is one of terms for concentration designation (along with percentage concentration, molar concentration, etc.). The size of a caption specifies what number of grams of substance contains in one milliliter of solution.
Instruction
1. Let's assume, such task is given. There are 20 milliliters of solution of sodium hydroxide. For its neutralization it was required to spend 30 milliliters 1M of solution of hydrochloric acid. Any of substances was not taken much. To define what caption of alkali.
2. First of all, write the reaction equation. It proceeds so: NaOH + HCl = NaCl + H2O.
3. You see that during this reaction of neutralization, according to the equation, the number of moles of acid completely coincides with the number of moles of the alkali connected by it. And how many moths of acid reacted? Time its solution one-molar, that number of moles will be in as much time less than unit how much 30 milliliters is less than 1 liter. That is, 30/1000 = 0.03 asking.
4. It follows from this that there were also 0.03 alkali asking. Consider how many it will be in grams. The molecular mass of a caustic natr is approximately equal 23 + 16 +1 = 40, therefore, its molar weight is 40 g/mol. Having increased 40 by 0.03, receive: 1.2 grams.
5. Well, and everything is farther very simply. 1.2 grams of alkali contain in 20 milliliters of solution. Having divided 1.2 into 20, receive the answer: 0.06 grams / milliliter. Here such caption at sodium hydroxide solution.
6. Let's complicate a statement of the problem. Let's assume, you have the same amount of solution of sodium hydroxide – 20 milliliters. For its neutralization added the same 30 milliliters 1M of hydrochloric acid. However, unlike the previous task, it turned out that acid was taken much, and for its neutralization it was necessary to spend 5 milliliters 2M of potassium hydroxide solution. What caption of solution of sodium hydroxide in this case?
7. Begin with writing of the equation of reaction of acid with caustic heat: HCl + KOH = KCl + H2O.
8. Arguing to similarly above-stated example and having made calculations, you will see: first, there were 0.03 initially hydrochloric acid asking, and secondly, with acid reacted 2х0.005 = 0.01 asking caustic heat. This alkali, respectively, connected 0.01 asking hydrochloric acid. Therefore, on the first reaction with other alkali – a caustic natr – left 0.03 - 0.01 = 0.02 asking hydrochloric acid. From what it becomes clear, as a caustic natr contained in solution 0.02 asking, that is, 40х0.02 = 0.8 grams.
9. And further there is no place to define a caption of this solution, in one action more simply. Having divided 0.8 into 20, receive the answer: 0.04 grams / milliliter. The solution of a task took slightly more time, but also there was nothing difficult.