Department
of Chemistry and Geosciences
Honors 1084 Lab Spring
2006
Lab
Test 7-Key
Name:_________________________________________________ Section:
Answer the following questions to the best of your ability showing all work necessary to obtain the answer. The answer must be given with the correct units and the correct number of significant figures where applicable. Remember to show all your work necessary to obtain the answer. Simply writing the correct answer without showing any work may not receive full credit.
1. (20 points) Why did you boil the solution after the indicator changed color the first time when you standardized the HCl solution with Na2CO3?
In order to remove the dissolved CO2 which
makes the indicator change color before the actual equivalence point.
2. (30 points) Using the same procedure as in last week’s experiment, 10.00 mL of a saturated Ca(OH)2 solution was placed in a clean 125 mL Erlenmeyer flask. About 25 mL of deionized water and 3 drops of bromothymol blue was added to the flask. The solution was then titrated to an endpoint with 0.01301 M HCl solution. It required 32.96 mL of the HCl solution to reach the endpoint. the neutralization reaction is:
2 HCl(aq) + Ca(OH)2(aq) ¾¾® CaCl2(aq) + 2 H2O(l)
(a) (5 points) How many moles of HCl were added to the reach the endpoint?
Calculate the number
of moles of HCl in 32.96 mL
of 0.01301 M HCl:
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(b) (5 points) How many moles of Ca(OH)2 were neutralized?
Using the acid-base
neutralization reaction above, convert moles of HCl
to moles of Ca(OH)2:
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(c) (10 points) What was the concentration of Ca(OH)2 in the saturated Ca(OH)2 solution?
Since 10.00 mL of Ca(OH)2 solution
was used, the concentration of Ca(OH)2 is equal to:

(d) (10 points) What is the value of Ksp for Ca(OH)2 from this information?
The Ksp expression for Ca(OH)2
is: Ksp
= [Ca2+][OH¯]2.
[Ca2+] is equal to
[Ca(OH)2] (= 0.02144 M) and [OH¯] = 2 ×
[Ca(OH)2] (= 0.04288 M). The Ksp value is:
Ksp = (0.02144)(0.04288)2 = 3.942 × 10-5
3. (30 points) In a standardization procedure, 0.2041 g of Na2CO3 was dissolved in 25 mL of deionized water and titrated against an HCl solution. If it required 25.67 mL of the HCl solution to reach the endpoint, what is the concentration of the HCl solution? The neutralization reaction is:
2 HCl(aq) + Na2CO3(aq) ¾¾® 2 NaCl(aq) + CO2(g) + H2O(l)
Calculate the number
of moles of Na2CO3:
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Convert to moles of HCl using the neutralization reaction shown above: 
Calculate the
concentration of HCl by dividing the number of moles
of HCl by the volume of HCl
solution added in liters (25.67 mL/(1000
mL/L) = 0.02567 L):
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4. (20 points) In today’s
experiment, you will be determining the amount of ______________ contained in a
sample of
nitrate ion
3. (20 points) Write the reaction associated with the Ksp for calcium hydroxide: