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in said method of measuring the volume of two joined together in reaction solutions, one of which - analyzed, and the second - titrant or volumetric solution - with a known concentration.For there is the concept of conditional titrant titer or a titer of the analyte.This amount of analyte titrate 1 ml of solution.The training course of chemistry, there are several types of problems in the calculation of the titer of the solution.
In the first type of tasks you will need to translate the concentration of the solution of the other units in the title.Concentration - is the ratio of the solute, predetermined mass, moles, v
olume, to the amount of the solution or solvent.In the solution based on the fact that to determine the titer of the original data must receive the weight of the solute and the size of volume of the solution in which it is located.
Example 1: Determine the titer of 15% sulfuric acid solution.Solution density 1.10 g / ml.Solution concentration is expressed in a mass fraction of substance.Mass fraction - the ratio of the masses of solute and solution.Calculate the weight of one liter of solution - 1100 grams.Determine the weight content of sulfuric acid in it 1100 * 0.15 = 165g.Calculate titer of the solution: 165 g / 1000 ml = 0.165 g / ml.
Example 2: you want to find a titer of 0.15 n.sulfuric acid solution.The normality of the solution - the number of equivalents of solute per liter of solution, the unit - mol-eq / l.Equivalent to - this quantity of substance equivalent to 1 mol of hydrogen ions in chemical reactions.One liter of a solution containing 0.15 mole equivalent sulfuric acid.
Using the periodic table, find the molar mass of H2SO4 - 98 g / mol.Equivalent of sulfuric acid - 1/2.Calculate the molar mass equivalent H2SO4: 98/2 = 49 g / mol.Find how much weigh 0.15 mol equivalents of sulfuric acid: 0.15 * 49 = 7.35 g Determine the titre of the solution: 7.36 g / 1000ml = 0.00736 g / ml.
In the second type of tasks you need to find a conditional title.To calculate the solutions of the initial values and the mass of solute the volume of the solution with which it has reacted.
Example 3: Calculate the titer of 0.1 n.AgNO3 solution for NaCl.Equivalents AgNO3 and NaCl are unity.Find the molar mass of NaCl - 58,5g / mol.Find the number of silver nitrate in 1 ml - 0.0001 mole.Therefore, the amount of sodium chloride, reacted with 1 ml - 0.0001 mole.Multiply the molar mass of NaCl per amount of substance and received a suspended titer of silver nitrate solution - 0.000585 g / ml - weight NaCl, 1 ml of solution reacts with AgNO3.
third type of problems - in the calculation of the titer of the solution from the values obtained titrimetrically.For their decisions are based on the analysis of the reaction equation.Get out of it, in which proportion the substance react with one another.
Example 4: Determine the titer of solution HCl, if the neutralization of 20 ml of acid required 18 ml of 0.13 n.NaOH solution.Equivalents of HCl and NaOH are equal to unity.Find the amount of sodium chloride in 18 ml: 0.13 * 0.018 = 0.00234 mole.Therefore, the amount of hydrochloric acid reacted is too 0.00234 mol.Calculate the molar mass HCl - 36,5 g / mol.Find the mass of the resulting amount of hydrochloric acid: 36.5 * 0.00234 = 0.08541 This mass of material contained in a 20 ml solution.Find a titer solution: 0.08541 / 20 = 0.0042705 g / ml.
- how to calculate the amount of the chemical