A titration is the laboratory technique of adding one solution to another.
A titration can be used to find an unknown concentration of sodium chloride in a solution by adding known volumes of the indicator sodium thiosulfate, sulfuric acid, and measuring the volume of hydrogen gas collected over an indicator.
The equivalence point in a titration is when both reactants are completely reacted and there are no more products being created.
The equivalence point will be reached when the difference between volumes for indicator added and volatile acid collected falls below 20%.
What quantity of nag3 was required to reach the equivalence point in the titration?
1. The equivalence point in the titration required quantity of nag3: 45 mL
To find the equivalence point in a titration, we must differentiate two equations, starting with the formation of a solution which is given by 1N NaOH = 1.00 mL (Ksp = 9.1 x 10) and ending with a product that is given by 45 mL of HCl = 50.0 grams (Ksp = 0.001).
First, all three terms on the right side of this equation are written as subscripts and grouped together as Na+(aq). Then, the charge of Na+(aq) is subtracted from H2O.
2. Establishing this relationship between the two equations is not clear cut.
Begin with the equivalence point as given by 1N NaOH = 1.00 mL (Ksp = 9.1 x 10) and end with a product that is given by 45 mL of HCl = 50.0 grams (Ksp = 0.001).
This equation contains three terms, Na+(aq), H2O, and HCl(aq). In Na+(aq) we are taken out of stock equilibrant and are at a potential to produce more than one product while in the other two terms, hydroxide and chlorine gas, we are in stock equilibrant.
The dissolved hydroxide and the hydrogen gas produced are both products of the titration.
The equivalence point is when both reactants are completely reacted and there are no more products being created.
The equivalence point will be reached when the difference between volumes for indicator added and volatile acid collected falls below 20%.
3. Write the equations for the titration –
NaOH(aq) + Na2S2O3(aq) acid –> Na+(aq) + S2O32-(aq)
Write the equations for the titration – Na+(aq) + HCl(aq) acid –> HCl(g) + Na+(aq)
Use these equations to find the result of adding 15 mL of NNaOH to 50 mL of HCl at the equivalence point.
For this problem, we have an end equation that we are trying to reach and a series of two different reactions that are leading up to that end result.
The equivalence point for this experiment is reached when the difference between the volumes measured for each component of the mixture (in mL) are less than 20% of one another.
If this difference is greater than 20%, that is not an equivalence point. To be a true equivalence point, both volumes must equal 0, but we’re trying to get as close to 0 as possible.
The volume of HCl(g) that we want to find is given by 50 g which, when divided by Na+(aq), gives us 5 mL. Divide half that value over 5 mL by our starting amount of 5 mL and we end up with 13.33%. Add 13.33% to 45 and we get 1.
4. Write the resulting equation –
2 Na+(aq) + S2O32-(aq) –> 2 NaS(s) + 2 SO42-(aq)
In an experiment that involves the titration of a solution, there will be a number of different reactions that take place between the two components in the solution.
Different things can happen at different points along this process. In each case, we must determine whether or not it is a neutralization reaction or an equivalence point that we are reaching.
To answer this question, all reactions must be written using their strengths, the amount of hydrogen gas produced relative to all the sodium thiosulfate produced in all reactions.
The equivalence point is reached when this difference falls below 20% of one another. For specific values of hydrogen gas, we can determine whether or not the reaction is a neutralization reaction by looking at the meanings for Ksp.
If the more hydrogen gas produced, the less Ksp , then it is a neutralization reaction. If Ksp is very small (lower than the equivalence point), then it is an equivalence point that has been reached.
5. Find the equivalence point –
NaOH(aq) + Na2S2O3(aq) acid –> Na+(aq) + S2O32-(aq)
The final term of this equation is hydroxide that was produced in a neutralization reaction. If we look at Ksp, we see that it is much less than the equivalence point.
In other words, this same relationship is true for hydrogen gas produced and hydroxide ions, which means to say that the relationship between these two components of this reaction are not exactly the same as the relationship between those of another reaction or portion of a reaction.
This means that the end equation (1N NaOH = 1.00 mL (Ksp = 9.1 x 10) and ending with a product that is given by 45 mL of HCl = 50.0 grams (Ksp = 0.001)) is not useful to find the equivalence point in this titration.
