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1) A 10 mL sample of HCl solution was transferred by pipet to an Erlenmeyer flask and then diluted by adding about 40 mL of distelled water. What is the approximate H3O+ concentration and pH of the solution in the flask before the titration begins? (approx concentration of HCl is 0.5 M)

2) Phenolphthalein indicator was added, and the solution in the flask was titrated with 0.215 M NaOH until the indicator turned pink (pH = 8-9). The exact volume of NaOH required was 22.75 mL. Use the following equation to calculate the concentration of HCl in the original 10 mL sample ==>

Mb x Vb = nMa x Va

Mb = molarity of standadrd base solution
Vb = volume of base added
n = mole ratio
Ma = unknown molarity of acid solution
Va = initial volume of acid solution

3) One student accidentally "overshot" the endpoint and added 23.90 mL of 0.215 M NaOH. Is the calculated concentration of HCl likely to be too high or too low as a result of this error? Why?

please help ASAP! thanks so much!

2007-10-03 09:24:52 · 1 answers · asked by KatieLC91 1 in Science & Mathematics Chemistry

1 answers

I think that you have a lot to catch up --- please work hard (on chem).
1) this sample got diluted 5 times, thus the approximate new concentration of HCl or H3O+ is 0.1 M (since HCl ionize to H3O+ and Cl-), and pH ≈ 1
2) 22.75 mL of 0.215 M NaOH is 0.02275*0.215 = 0.004891 Mol of NaOH, which may completely neutralize 0.004891 Mol of HCl. So the original molar concentration in 10ml HCl is: 0.004891/0.01 = 0.489M.
3) The relative error is (23.90 - 22.75)/22.75 = 1.15/22.75 = 5.05%
The calculated concentration of HCl likely to be 5.05% high, since the concentration calculation of HCl is based on molar mass of NaOH used.

2007-10-03 11:55:23 · answer #1 · answered by Hahaha 7 · 3 0

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