We performed an experiment where we got 25mL of standard 2.0M NaOH, then used a

Question

We performed an experiment where we got 25mL of standard 2.0M NaOH, then used a 10mL pipet to transfer one aliquot of it to a 100mL volumetric flask and then diluted to the mark on the flask with distilled water.

We then got acetic acid and pipetted with a 5ml pipet into a 150mL beaker, and then added 70mL of distilled water to the beaker.

We then filled a buret with the diluted standard NaOH and then started adding this to the beaker with the acetic acid in small amounts while checking the pH of the acetic acid. We did this until we hit the equivalence point and the pH started rising quickly. We did all of this 3 times total.

I am confused about some questions on this experiment.

1) Is the concnetration of the standard solution simply 0.5M, because we do M2 = M1V1 / V2 where M1 = 2.0M, V1 = 25mL, and V2 = 100mL?

2) How do I find the concentration of the acetic acid for each of the three trials?

Explanation / Answer

1) The molarity of added NaOH:

M1V1 = M2V2

i.e. 2 M * 10 mL = M2 * 100 mL

i.e. M2 = 20/100 = 0.2 M

i.e. Mb = molarity of NaOH = 0.2 M and Vb = volume of NaOH = 100 mL

2) Here, NaOH solution is added to the acetic acid solution until the equivalence point:

i.e. MaVa = MbVb, where Ma is the molarity of acetic acid and Va is the volume of acetic acid.

i.e. Ma * 75 mL = 0.2 M * 100 mL

i.e. Ma = 20/75 = 0.267 M

Therefore, the concenration of acetic acid in each trial = 0.267 M

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