You have sonic NaOH remaining in the buret at the end of part A. which you finis

Question

You have sonic NaOH remaining in the buret at the end of part A. which you finished one hour ago. Not wanting to be wasteful, you add this solution back into tin- original container. Will this affect the concentration of the NaOH If so, why? You use unboiled DI-H_2O to dissolve your potassium hydrogen phthalate sample and you calculate NaOH to be 0.155 m. (a) NaOH reacts with CO_2. Show this reaction. Under normal conditions, carbon dioxide is present in water at around 1.20 times 10^-5 M. Taking into account the fact that some NaOH reacted with CO_2 instead of potassium hydrogen phthalate, what is your actual NaOH molarity? (c) Is this a large enough difference to notice using our precision glassware and balances?

Explanation / Answer

Ans. 1. NaOH is hygroscopic, that is absorbs moisture from atmosphere. An aqueous solution of NaOH is also hygroscopic. When left in a burette, which is open at top making the solution in direct contact of atmospheric acid, NaOH solution is most likely to absorbs moisture from air and increase its volume. Increase in volume (while amount of NaOH remains unaffected) dilutes the solution.

So, if diluted NaOH is transferred from burette to original container, the final solution all gets diluted.

Note: the extent of dilution depends on the strength of NaOH (greater is molarity, more hygroscopic is solution), relative humidity in air, etc.

Ans. 2. A. Balanced Reaction:          NaOH(aq) + CO2(g) ----------> NaHCO3(aq)

Ans. 2. B. In a solution containing CO2 dissolved in it, the moles of NaOH wasted to react with CO2 is equal to moles of CO2 dissolved in solution.

Molarity of NaOH = 0.155 M

Molarity of CO2 = 1.20 x 10-5 M = 0.0000120 M

Assuming 1.0 L solution, net NaOH concentration after CO2 reaction is-

            0.155 M – 0.0000120 M = 0.154988 M

Ans. 2. C. Error in NaOH molarity = Initial [NaOH] – Actual [NaOH] after CO2 reaction

                                                = 0.155M – 0.154988 M = 0.004988 M

% error = (Error in molarity/ Theoretical molarity) x 100

                        = (0.004988 M / 0.155 M) x 100

                        = 3.218 %

The % error is considerably high and must be accounted, theoretically.

However, the difference in molarity, volume of solution and mass of solution due to reaction of NaOH with CO2 is most likely to go un-noticed by precision glassware and balances. Because glassware can read volume in two decimal point and precision balance in three decimals- in both cases the last decimal point being uncertain. That is to notice difference between 0.155 and 0.154988, there must a system (say, analytical balance) that reads in at least four decimal point, so that the third decimal point must be correctly noted. None of the precision glassware and balance can notice such small (third decimal) difference.

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