AcidSolutionABCDVolume ofNaOH Added (mL)4075115200
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
titrated with a 0.20 M NaOH solution. The volume of NaOH needed to reach the end point for each sample is given
in the table above. Which solution is the most suitable to maximize the yield?
To reach end point, the number of moles of hydronium ions and hydroxide ions are equivalent.
We first calculate the range of hydronium ions available.
Since the acid is monoprotic, the moles of acid is equal to the moles of hydronium:
Lower value
0.20M Acid=1 L0.20 mol
1 L0.20 mol×0.100 L=0.02 mol
Upper value
0.30M Acid=1 L0.30 mol
1 L0.30 mol×0.100 L=0.03 mol
We can do a similar calculation for the base to see how many hydoxide ions are present.
Solution A
0.20M Hydroxide=1 L0.20 mol
1 L0.20 mol×0.040 L=0.008 mol
Solution C
0.20M Hydroxide=1 L0.20 mol
1 L0.20 mol×0.115 L=0.023 mol
Solution B
0.20M Hydroxide=1 L0.20 mol
1 L0.20 mol×0.075 L=0.015 mol
Solution D
0.20M Hydroxide=1 L0.20 mol
1 L0.20 mol×0.200 L=0.040 mol
Solution C provides an end point in which the number of moles of hydroxide ions is equal to an acceptable number of moles of hydronium ions.