Need answers to 1C and 2, 3!! 1. Record your results for each of the 3 trials: 2

Question

Need answers to 1C and 2, 3!!

1. Record your results for each of the 3 trials:

2. Calculate the concentration of the CoCl2 solution (moles/L)

3. What is the average concentration of the CoCl2 solution? To how many significant digits can this concentration be reported? (Consider the accuracy of the burette and the volume in one drop.)

The concentration of the cobalt(II) chloride solution is determined by comparing the concentrations of the titrant and titrate. The equation for the chemical reaction between cobalt chloride and sodium hydroxide is:

CoCl2(aq) + 2NaOH (aq) 2NaCl (aq) + Co(OH)2(s)

At the endpoint, exactly enough NaOH has been added to remove the cobalt ions from the solution in the form of the solid precipitate, Co(OH)2. Notice, however, that 2 molecules of NaOH are required to react with one molecule of CoCl2. Therefore, if we know the total moles of NaOH used in the titration, then there will have been half as many moles of CoCl2in the titrated sample.

This means that we must adjust the familiar formula for a titration to account for the ratio in which the CoCl2 and NaOH react:

(Moles of CoCl2 ) = (Moles of NaOH) ÷ 2

which can be expressed in terms of concentrations as:

C1 Ý V1 = (C2 Ý V2) ÷ 2

where:

a Volume of CoCl2 (mL)   10.00
  
10.00
10.00 b Volume of NaOH added (mL) 18.48
   18.04
  
18.01 c Concentration of NaOH added

Explanation / Answer

with out NaOH concentration we can not answer the questions. If you consider the concentration of NaOH is 1 moles/L then answers are as follow

1c) 1 moles/L in all cases

2) from the formula C1V1=2C2V2

C1 X 10.0 = 2 X 1X 18.48

= 3.696 moles/L

similarly for the other values C1 will be 3.608 and 3.602 moles/L

mean of the three values = concentration of CoCl2 = (3.696+3.608+3.602)/3

=3.653 moles/L

standard deviation is [(1/3) X [(3.696-3.653)2+(3.608-3.653)2+(3.602-3.653)2]]1/2

   = 0.046

so we can report concentration upto first digit (3.6 moles/L)

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