In the titration of the unknown solid acid, you stirred the mixture too vigorous
ID: 510755 • Letter: I
Question
In the titration of the unknown solid acid, you stirred the mixture too vigorously when dissolving in water and some of the unknown acid solution splashes out of the container before you started the titration,
a) How is the titration volume of the titrant, NaOH affected-too much, too little, same amount? Explain.
b) How will the gram equivalent weight of the unkown solid acid be affected: too high, too low, or no effect? Explain.
c) Why would the gram equivalent weight of the solid acid be calculated instead of the molar mass?
d) How does calculating the ratio of gram of acid to mL NaOH titrated help you to evaluate the precision of the titration?
Explanation / Answer
a) The titration volume of the titrant, NaOH will be affected too much. This is because, there will be considerable decrease in the volume of the acid solution on splashing.
b) There will be no effect on the gram equivalent weight of the unknown solid acid because, splashing of the solution causes a decrease in volume but not the concentration. The gram equivalent weight is calculated from the Normality (concentration) of the unknown solid acid solution using the formula V1N1 = V2N2 . You can understand the argument from the formula itself. V2 (titrant volume) decreases as V1 (acid solution volume) decreases on splashing. But N1 (concentration of the acid solution) is not affected.
c) Volumetric titration calculations are done using the formula V1N1 = V2N2 .The formula is based on the fact that the number of equivalents of the acid = number of equivalents of the base at the equivalence point and not the number of moles. So, we cannot calculate molar mass. From the titration, you can calculate the Normality of the unknown acid solution using the above formula. As you know the weight of the solid acid you have initially added to water, you can calculate the gram equivalent weight.
Normality = Number of equivalents per L.
Number of equivalents = weight/equivalent weight.
d) N1/V2 = N2/V1 After calculating N1, you can evaluate the precision of the titration by comparing these two ratios.The ratio, N1/V2 is known accurately. N1/V2 is nothing but the ratio of gram of acid/equivalent weight of acid to volume of NaOH. In this, equivalent weight is not variable. So, you can consider N1/V2 as the ratio of gram of acid to mL NaOH. The extent to which, the ratio N2/V1 differs from the ratio N1/V2 will tell you about the loss in precision because of splashing. You do not know the actual volume after splashing. You know only the volume of water initially taken in which a known weight of the solid acid was dissolved, which you are using as V1 in your calculations.