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Chapter 12: Q3 TY (page 275)
Find if free, unprotonated [NH3] = 0.02 M.
=0.00673 if free, unprotonated [NH3] = 0.02 M.
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Spreadsheet equation for auxiliary complexing agent. Consider the titration of metal M (initial concentration = CM, initial volume = VM) with EDTA (concentration = CEDTA, volume added = VEDTA) in the presence of an auxiliary complexing ligand (such as ammonia). Follow the derivation in Section 12-4 to show that the master equation for the titration is
where role="math"> is the conditional formation constant in the presence of auxiliary complexing agent at the fixed pH of the titration (Equation 12-18) and [M]free is the total concentration of metal not bound to EDTA. [M]free is the same as [M] in Equation 12-15. The result is equivalent to Equation 12-11, with [M] replaced by [M]free and Kf replaced by .
Indirect EDTA determination of cesium. Cesium ion does not form a strong EDTA complex, but it can be analyzed by adding a known excess of NaBiI4 in cold concentrated acetic acid containing excess NaI. Solid Cs3Bi2I9 is precipitated, filtered, and removed. The excess yellow is then titrated with EDTA. The end point occurs when the yellow color disappears. (Sodium thiosulfate is used in the reaction to prevent the liberated from being oxidized to yellow aqueous I2 by O2 from the air.) The precipitation is fairly selective for Cs+. The ions Li+, Na+, K+, and low concentrations of Rb+ do not interfere, although Tl+ does. Suppose that 25.00 mL of unknown containing Cs+ were treated with 25.00 mL of 0.08640 M NaBiI4 and the unreacted required 14.24 mL of 0.0437 M EDTA for complete titration. Find the concentration of Cs+ in the unknown.
Calculate pCu2+ (to the 0.01 decimal place) at each of the following points in the titration of 50.0 mL of 0.040 0 M EDTA with 0.080 0 M Cu (NO3)2 at pH 5.00: 0.1, 5.0, 10.0, 15.0, 20.0, 24.0, 25.0, 26.0, and 30.0 mL. Make a graph of pCu2+ versus volume of titrant.
Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn2+ at the following volumes of added EDTA and sketch the titration curve:
(a) 0 mL (b) 20.0 mL (c) 40.0 mL (d) 49.0 mL (e) 49.9 mL (f) 50.0 mL (g) 50.1 mL
(h) 55.0 mL (i) 60.0 mL
A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn2+ will be required for back titration?
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