In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the

Question

In our lab experiment, we filtered CaCl and CaOH2 solutions that we prepared the previous week and transferred both of the filtrates using a 25 ml pipette into two erlenmeyer flasks. We then prepared the standard HCl by pipeting 100 ml of HCl to prepare 250 ml of 0.04M HCl from the 0.1 M HCl solution that we initially had. We pipeted 100 ml of HCl into a 250 ml volumetric flask and diluted to the mark. The buret was filled with the prepared HCl and both the CaCl and CaOH2 were titrated until a yellow endpoint was reached, this titration was repeated three times for each solution. The data that I obtained was:

CaCl= Initial(ml)--Final (ml) --- Volume of HCl(ml)

1.30 --24.2--22.9

0.50 --23.5 --23.0

1.60 --23.2 -- 21.6

Ca(OH)2= initial--final-- volume of HCl

1.00-- 28.3-- 27.3

2.00-- 28.8-- 26.8

2.00-- 28.1 --26.1   

Q1. For each sample titrated, calculate the mean and standard deviation of your two or more titration volumes.

Q2. Using mean titration volumes, calculate the hydroxide concentration in each of the two saturated Ca(OH)2 solutions.

Explanation / Answer

1) Note down the mean titration volumes for titration of CaCl2: 22.9 mL, 23.0 mL and 21.6 mL.

Mean of the titration volumes = (22.9 + 23.0 + 21.6) mL/3 = 22.5 mL (ans).

Standard deviation of the mean = [(22.9 – 22.5)2 + (23.0 – 22.5)2 + (21.6 – 22.5)2]/3 = 1.22/3 = 0.4067 = 0.6378 0.64 (ans).

Next, consider the titration of Ca(OH)2. The titration volumes are 27.3 mL, 26.8 mL and 26.1 mL.

Mean of the titration volumes = 1/3*(27.3 + 26.8 + 26.1) mL = 26.73 mL 26.7 mL (ans).

Standard deviation of the mean = [(27.3 – 26.7)2 + (26.8 – 26.7)2 + (26.1 – 26.7)2]/3 = 0.4933 0.49 (ans).

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