Homework Answers
a) Minimum stability constant required to titrate 0.01 M
Cd2+ with 0.01 M EDTA = Kf x
Where Kf is the stability constant = 2.9 x
1016, = 0.37 at pH
10
Minimum stability constant at pH 10 = 1.07 x 1016
b) Minimum pH for titrating 0.01 M Cd2+ with 0.01 M EDTA is 10.17. Below this pH sufficient amount of EDTA4- species are not available to form the complex
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What is the minimum formation constant required to titrate 0.010 M Cd2 with 0.010 M EDTA,...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7 mL of 0.0500M EDTA....
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7
mL of 0.0500M EDTA. Titration of the excess unreacted EDTA required
15.8 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the
addition of an excess of CN−. Titration of the newly freed EDTA
required 13.7 mL of 0.0190 M Ca2+. What are the concentrations of
Cd2+ and Mn2+ in the original solution?
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7...
Answer all questions You are titrating 20.0 mLs of 0.10 M Cd2+ with 0.20 M EDTA at pH 9.0, using 1.0 M acetate to pr...
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You are titrating 20.0 mLs of 0.10 M Cd2+ with 0.20 M EDTA at pH 9.0, using 1.0 M acetate to prevent Cd2+ from precipitating as Ca(OH)2. Calculate the concentration of uncomplexed Cd2+ for the following volumes of added EDTA. a. 0 mLs b. 8.0 mLs c. 10.0 mLs d. 11.0 mLs
A sample containing Cd2+ and Mn2+ was treated with 60.1 ml of 0.0400M EDTA. Titration of...
A sample containing Cd2+ and Mn2+ was treated with 60.1 ml of 0.0400M EDTA. Titration of the excess unreacted EDTA required 16.8ml of 0.013M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN-. Titration of the newly freed EDTA required 10.4ml of 0.013M Ca2=. What are the concentrations of Cd2+ and Mn2+ in the original solution?
The formation constant for the lead-EDTA complex (PbY^2 -) is 1.10 times 10^18. At pH 10.0,...
The formation constant for the lead-EDTA complex (PbY^2 -) is 1.10 times 10^18. At pH 10.0, alpha is found to be 0.35. Calculate pPb^2+ for 50.00 mL of a solution of 0.200 M Pb^2+ at pH 10 after the addition of 150.00 mL of 0.010 M EDTA.
2 poin QUESTION 5 What is the value of conditional formation constant, Kr. for the EDTA-complex...
2 poin QUESTION 5 What is the value of conditional formation constant, Kr. for the EDTA-complex with Cu2+ at pH 10.07 Kr(Cuy) - 6.03x1018 Cu2+ + EDTA CuY2 (Use Table 11-1 in the textbook as a reference.) 1.2.01 x 1019 2.1.81 x 1018 3.3.21 x 1015 O4 7.66 x 10-20 5.6.03 x 1018 QUESTION 6 2 poir Consider titration of 24.00 ml of 0.0880 M Cuat solution with 0.110 M EDTA at pH 10.0. Find the concentration of free Cu2+...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN“. Titration of the newly freed EDTA required 11.3 mL of 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 42.5 mL of 0.0700 M...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 42.5 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 16.8 mL of 0.0120 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 26.9 mL of 0.0120 M Ca2+. What were the molarities of Cd2+ and Mn2+ in the original solution? Answer: ([Cd2+] = 0.00646 M; and [Mn2+] = 0.0490 M) The...
Calculate the value of the conditional formation constant, Ki', for EDTA with the following metal ions....
Calculate the value of the conditional formation constant, Ki', for EDTA with the following metal ions. State for each if a titration of that ion at the listed pH would be quantitative or not. (Consider the pH values as exact numbers.) a. K+ at pH 8 Ca2+ at pH 11 Co3+ at pH 1 Cu2+ at pH 4
Please help! Will rate. A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5...
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
You are titrating 130.0 mL of 0.050 M Ca2 with 0.050 M EDTA at pH 9.00....
You are titrating 130.0 mL of 0.050 M Ca2 with 0.050 M EDTA at pH 9.00. Log K for the Ca2 EDTA complex is 10.65, and the fraction of free EDTA in the Y4-form, *. is 0.041 at pH 9.00. (a) What is Kt, the conditional formation constant, for Ca2 at pH 9.00? Number K: (pH 9.00) = 110 (b) What is the equivalence volume, Ve, in milliliters? Number mL (c) Calculate the concentration of Ca+ at V- 1/2 Ve...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7
mL of 0.0500M EDTA. Titration of the excess unreacted EDTA required
15.8 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the
addition of an excess of CN−. Titration of the newly freed EDTA
required 13.7 mL of 0.0190 M Ca2+. What are the concentrations of
Cd2+ and Mn2+ in the original solution?
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7...
Answer all questions
You are titrating 20.0 mLs of 0.10 M Cd2+ with 0.20 M EDTA at pH 9.0, using 1.0 M acetate to prevent Cd2+ from precipitating as Ca(OH)2. Calculate the concentration of uncomplexed Cd2+ for the following volumes of added EDTA. a. 0 mLs b. 8.0 mLs c. 10.0 mLs d. 11.0 mLs
The formation constant for the lead-EDTA complex (PbY^2 -) is 1.10 times 10^18. At pH 10.0, alpha is found to be 0.35. Calculate pPb^2+ for 50.00 mL of a solution of 0.200 M Pb^2+ at pH 10 after the addition of 150.00 mL of 0.010 M EDTA.
2 poin QUESTION 5 What is the value of conditional formation constant, Kr. for the EDTA-complex with Cu2+ at pH 10.07 Kr(Cuy) - 6.03x1018 Cu2+ + EDTA CuY2 (Use Table 11-1 in the textbook as a reference.) 1.2.01 x 1019 2.1.81 x 1018 3.3.21 x 1015 O4 7.66 x 10-20 5.6.03 x 1018 QUESTION 6 2 poir Consider titration of 24.00 ml of 0.0880 M Cuat solution with 0.110 M EDTA at pH 10.0. Find the concentration of free Cu2+...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN“. Titration of the newly freed EDTA required 11.3 mL of 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+
Calculate the value of the conditional formation constant, Ki', for EDTA with the following metal ions. State for each if a titration of that ion at the listed pH would be quantitative or not. (Consider the pH values as exact numbers.) a. K+ at pH 8 Ca2+ at pH 11 Co3+ at pH 1 Cu2+ at pH 4
Please help! Will rate.
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5 mL of 0.0600 M EDTA. Titration of the excess unreacted EDTA required 13.3 mL of 0.0100 M Ca. The Cd2t was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 11.2 mL of 0.0100 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? M Mn2+ concentration: M Cd2 concentration: A...
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