To maximize the yield in a certain manufacturing process, a solution of a weak monoprotic acid that has a concentration between
0.20 M and 0.30 M is required. Four 100. mL samples of the acid at different concentrations are each titratedwith a 0.20 M NaOH solution. The volume of NaOH needed to reach the end point for each sample is given in the table. Which solution is the most suitable to maximize the yield? Extra Content
Acid Solution Volume of NaOH added (mL)
A 40 mL
B 75 mL
C 115 mL
D 200 mL
A. Solution A
B. Solution B
C. Solution C
If you have a solution of a monoprotic acid, it means that it has the form HA (just one Hydrigen atom). Therefore, one molecule of acid is going to react with just one molecule of NaOH.
So, if you have a solution of 100 ml of 0,2 HA acid, it is going to react with a 100 ml of 0,2 M NaOH solution. Beacause we know that the acid can be a little more concentrated than 0,2 M (0,2-0,3), it probably needs a little more than 100 ml of NaOH to react. So, the answer is C.
in 100 ml of 0,2M acid you can find 0,002 mol of HA
in 100 ml of 0,3 M acid you can find 0,003 mol of HA
in 100 ml of 0,2 M NaOH you can find 0,002 mol of NaOH
the answer can not be 200 because in 200 ml of 0,2 M NaOH there are 0,004 mol of NaOH, which is more than 0,003 mol.