Question

A student pipets 5.00 mL of a 5.103 M aqueous NaOH solution into a 250.00 mL volumetric flask and dilutes up to the mark with distilled water. What is the final molarity of the dilute solution? Select one: A. 0.1 MX B. 0.10206 M C. 0.102 M D. 0.1021 M Your answer is incorrect. 86.14 mL of an acid solution was needed to neutralize 30.24 mL of a base solution of unknown concentrations. A second trial is run but this time 30.24 ml base solution is diluted to a total volume of 50.00 mL before starting the titration. How many mL of the acid solution are needed to neutralize it? Select one: A. 52.10 ml B. 142.42 mL C. 105.90 mL X O D. 45.26 mL E. 86.14 mL

Answer 1

1- The given initial concentration (M1) of NaOH = 5.103 M

given initial volume (V1) of NaOH = 5.00 ml

Given final volume (V2) of NaOH = 250.00 ml

Then the final concentration (M2) can be found from the formula- M1V1 = M2V1

Thus M2 = M1V1 / V2

= 5.103 M * 5.00 ml / 250.00 ml

= 0.10206 M

It shows that volume and concentration has an inverse relation. If we increase the volume by x times, then the concentration of the solution simultaneously decreases by x times.

2-

Given 86.14 ml acid solution is required to neutralize 30.24 ml of base solution. Here the concentration of base solution is unknown.

Lets consider we have X moles of base in the base solution.

Now given in 2nd titration, the base solution is diluted to 50 ml. Here the dilution is done by adding more water. That means in our new solution, only the total volume of solution increased. But there is no increase in the number of moles of base. The number of moles of base in the 2nd solution is still the same i.e = X moles

Lets say our initial acid solution has Y moles of acid. Thus Y mols of acid is capable of neutralizing X moles of base.

Since there is no change in the number of moles of base in the 2nd solution, thus there should be no change in the number of moles in the acid. Our acid should be = Y moles

And since the initial 86.14 ml of acid solution already has Y mols of acid, thus it can diretly be added to the base solution of 50 ml

So the required volume of acid solution = 86.14 ml