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Question: Explain the use of preparing standard solutions and titrations....

Explain the use of preparing standard solutions and titrations.

Explanation

Solution

Standard solution is a reagent of known concentration that is used to carry out a volumetric titration. The reagent must be stable, pure, and preferably of high molecular weight. Standard solution has many uses – to determine the concentration of the analyte, to find out the number of moles, etc.

Complete answer:
To prepare a standard solution, a known mass of solute is dissolved and the solution is diluted to a precise volume. Standard solution concentration is usually expressed in terms of molarity (M) or moles per liter(molL)\left( {\dfrac{{mol}}{L}} \right). Not all substances are suitable solutes for standard solutions.
There are many uses of standard solutions:
A standard solution is used to determine the concentration of the analyte during titration.
It is used to find out the number of moles in the solution in the reaction.
To understand it better let us take an example:
Suppose we make a standard solution of sodium carbonate where (13.25 g)\left( {13.25{\text{ }}g} \right) was dissolved in about 150 mL150{\text{ }}mL of deionized water in a beaker. The solution was transferred into a 250 mL250{\text{ }}mL volumetric flask and was made up to the mark. This is our standard solution.
A 25 mL25{\text{ }}mLvolume of the sodium carbonate solution was pipetted into a conical flask. Methyl orange indicator was added.
The sample required 24.65 mL24.65{\text{ }}mLof a hydrochloric acid solution to completely neutralize it.
Now if we have to calculate the mole and molarity of Na2CO3N{a_2}C{O_3}, it will be calculated in the following manner-
Moles of Na2CO3= 13.25 g Na2CO3×  1mol Na2CO3105.99g Na2CO3  = 0.12501 mol Na2CO3N{a_2}C{O_3} = {\text{ }}13.25{\text{ }}g{\text{ }}N{a_2}C{O_3} \times \;\dfrac{{1mol{\text{ }}N{a_2}C{O_3}}}{{105.99g{\text{ }}N{a_2}C{O_3}}}\; = {\text{ }}0.12501{\text{ }}mol{\text{ }}N{a_2}C{O_3}
And molarity of Na2CO3=Molarity =  moleslitres=0.12501mol0.2500L  = 0.500 05 mol/LN{a_2}C{O_3} = Molarity{\text{ }} = \;\dfrac{{moles}}{{litres}} = \dfrac{{0.12501mol}}{{0.2500L}}\; = {\text{ }}0.500{\text{ }}05{\text{ }}mol/L
Thus standard solutions are an important part of titrations.

Note:
It is important to note that the concentration of the standard solution must remain constant all the time. This is so that there is no need for standardization. Its reaction with the analyte must be rapid to minimize the waiting period after the addition of each reagent. It should be possible to describe the reaction by a balanced chemical reaction.