Question

Focus on Step A5 of the procedure, where one partner prepared their sample for visual titration by adding approximately 100mL of water to the to sample in the plastic cup or beaker before beginning the titration.

Note that this step was also repeated in the potentiometric (pH-meter) titration in Part 2 (Step A7).

In both cases, the addition of more water was an experimental necessity, to allow the solution to be stirred easily and to ensure that the pH probe is fully immersed in the solution (to obtain accurate pH readings during the titration.

• It seems strange that we can be so casual about adding an approximate volume here, when all of our prior measurements in the experiment were so precisely completed using volumetric glassware.

To understand why this is reasonable, we will look at what effects this added water had on the titration data collected. This was originally a margin question in the Experiment 4 procedure (Page 9).

Let's imagine what would happen if we repeated the titration without adding 100 mL of extra RO water to the plastic cup at the start. Predict if (and how) the titration would be changed if we skipped this step.

Compared to the actual titration performed in the laboratory, during this imaginary new titration...

Question 1 options:

... the volume of NaOH needed to reach equivalence would be... ... the concentration of conjugate weak base at equivalence would be... ... the pH at the half-equivalence point would be... ... the pH at the start of the titration (before any NaOH is added) would be...

1. higher 2. lower 3. the same

Answer 1

Solution: Margin Question: why we do not need to precisely measure the volume of water added here? Answer This because that addition of extre water do not change the moles of acid present in plashic cup. That is moles of Acid before addition of water is same as after addition of water. But the concentration of acid changes since volume of solution increases, we can use the dilution formula to calculate the molarity of auid after addition of loome water. mivi - Mavz before addition → Questions after addition . product of molosity and volumes gives number 8 Moles. @ volume of Naon neacted to the reach the equivalence point would be (same] because moles of aciet is same hence equal moles will be required to of Naon will be required to react with acid hence equal volume will be required. Ics Scanned with * CamScanner

B concentration of conjugane weak base at equivalence would be higher because maly of conjugale weako base formed at equiredence point will be same in both cases but in this case . volume is less. so concentration will be high. . molarity = no. of moles 0 volume modersity of volume ® pr at half equivalence point. (same] since we are titrating a weak and with Naon so it will form a buffer solution and it half equivalence point [aud] =. . [ conjugare base and pH = pka bah, ( р. ehne j4 6 Ио ри bе и Es Scanned with CamScanner

e pH at start of titration : lower .ph of weak acid ph depends on the concentration of [ut] High concentration of nt means low ph. if we will not add toome H2o then concentration of H+ ion is solution will be high i.e. low pri then cs Scanned with CamScanner