Exam 13: Edta Titrations

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EDTA is a hexadentate ligand containing four carboxylic acid groups and two amines.Which is true for the acid-base properties of EDTA? I EDTA is a hexaprotic system,H6Y2+,when the amines are protonated. II Neutral EDTA,H4Y,is tetraprotic when the amines are not protonated. III Deprotonated EDTA,Y4−,is the only form of EDTA that binds to metal cations. IV The fraction of EDTA in any of the six EDTA species is pH dependent.

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E

Which is NOT an end point detection method used with EDTA titrations?

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B

A____________________ titration is necessary for analytes that precipitates in the absence of EDTA,that reacts slowly with EDTA or that blocks the indicator.

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E

Calculate pBa when 35.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+.For the buffered pH of 10, Calculate pBa when 35.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba<sup>2+</sup>.For the buffered pH of 10,   = 0.30.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 0.30.Kf = 7.59 × 107 for BaY2−.

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Calculate [Mn2+] when 50.00 mL of 0.1000 M Mn2+ is titrated with 17.00 mL of 0.2000 M EDTA.The titration is buffered to pH 11 for which Calculate [Mn<sup>2+</sup>] when 50.00 mL of 0.1000 M Mn<sup>2+</sup> is titrated with 17.00 mL of 0.2000 M EDTA.The titration is buffered to pH 11 for which   = 0.81.K<sub>f</sub> = 7.76 × 10<sup>13</sup> = 0.81.Kf = 7.76 × 1013

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Which statement is NOT true?

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Calculate Calculate   for a 0.2 M EDTA solution buffered at pH 10.00.For EDTA K<sub>1</sub> = 1.0,K<sub>2</sub> = 31.6,K<sub>3</sub> = 100,K<sub>4</sub> = 490,K<sub>5</sub> = 1.35 × 10<sup>6</sup> and K<sub>6</sub> = 2.34 × 10<sup>10</sup>. for a 0.2 M EDTA solution buffered at pH 10.00.For EDTA K1 = 1.0,K2 = 31.6,K3 = 100,K4 = 490,K5 = 1.35 × 106 and K6 = 2.34 × 1010.

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Calculate [Ba2+] when 35.00 mL of 0.2000 M EDTA is added to 50.00 mL of 0.1000 M Ba2+ in the presence of 0.1 M nitrilotriacetate. Calculate [Ba<sup>2+</sup>] when 35.00 mL of 0.2000 M EDTA is added to 50.00 mL of 0.1000 M Ba<sup>2+</sup> in the presence of 0.1 M nitrilotriacetate.   = 1.48 x 10<sup>−</sup><sup>4</sup>.   = 0.30 for pH 10.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 1.48 x 10−4. Calculate [Ba<sup>2+</sup>] when 35.00 mL of 0.2000 M EDTA is added to 50.00 mL of 0.1000 M Ba<sup>2+</sup> in the presence of 0.1 M nitrilotriacetate.   = 1.48 x 10<sup>−</sup><sup>4</sup>.   = 0.30 for pH 10.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 0.30 for pH 10.Kf = 7.59 × 107 for BaY2−.

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The reaction between Al3+ and EDTA is too slow for direct titration.To a 25.00 mL Al3+ sample 50.00 mL of 0.2000 M EDTA is added and allowed to react for 30 minutes.The excess EDTA was titrated with 21.23 mL of 0.1000 M Zn2+.What is the concentration of Al3+ in the sample?

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Calculate pBa when 75.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+.For the buffered pH of 10, Calculate pBa when 75.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba<sup>2+</sup>.For the buffered pH of 10,   = 0.30.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 0.30.Kf = 7.59 × 107 for BaY2−.

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Calculate [Mn2+] when 50.00 mL of 0.1000 M Mn2+ is titrated with 25.00 mL of 0.2000 M EDTA.The titration is buffered to pH 11 for which Calculate [Mn<sup>2+</sup>] when 50.00 mL of 0.1000 M Mn<sup>2+</sup> is titrated with 25.00 mL of 0.2000 M EDTA.The titration is buffered to pH 11 for which   = 0.81.K<sub>f</sub> = 7.76 × 10<sup>13</sup> = 0.81.Kf = 7.76 × 1013

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The____________________ is the ability of multidentate ligands to form more stable metal complexes than those formed by similar monodentate ligands.

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A____________________ is a reagent that protects a component of the analyte from reaction with EDTA.

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Which statement is NOT true for auxiliary complexing agents?

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A metal that does not freely dissociate from an indicator is said to____________________ the indicator.

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Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+ in the presence of 0.1 M nitrilotriacetate. Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba<sup>2+</sup> in the presence of 0.1 M nitrilotriacetate.   = 1.48 × 10<sup>−</sup><sup>4</sup>.   = 0.30 for pH 10.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 1.48 × 10−4. Calculate pBa when 35.00 mL of 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba<sup>2+</sup> in the presence of 0.1 M nitrilotriacetate.   = 1.48 × 10<sup>−</sup><sup>4</sup>.   = 0.30 for pH 10.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 0.30 for pH 10.Kf = 7.59 × 107 for BaY2−.

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For EDTA titrations the titration reaction is Mn+ + EDTA ⇋ MYn-4.Which statements are true for EDTA titration curves? I At the equivalence point,there is exactly as much EDTA in solution as metal.[Mn+] is calculated from the dissociation of MYn-4 formed. II After the equivalence point,the concentration free EDTA equals excess EDTA and virtually all metal ion is in the form MYn-4.[Mn+] is calculated from the dissociation of MYn-4. III Before the equivalence point,free [Mn+] equals excess unreacted Mn+ after EDTA has been consumed.Dissociation of MYn-4 is negligible. IV pM is plotted on the y-axis and milliliters EDTA solution added on the x-axis.

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Which statement is incorrect for conditional formation constants,Kf′?

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Ammonia,0.10 M,is used as an auxiliary complexing agent for a Co2+ titration with EDTA.There are 6 s for the complex between ammonia and Co2+.Which equation is used to calculate Ammonia,0.10 M,is used as an auxiliary complexing agent for a Co<sup>2+</sup> titration with EDTA.There are 6 <font face=symbol></font>s for the complex between ammonia and Co<sup>2+</sup>.Which equation is used to calculate   ? ?

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Calculate pBa when 50.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba2+.For the buffered pH of 10, Calculate pBa when 50.00 mL 0.1 M EDTA is added to 50.00 mL of 0.1 M Ba<sup>2+</sup>.For the buffered pH of 10,   = 0.30.K<sub>f</sub> = 7.59 × 10<sup>7</sup> for BaY<sup>2</sup><sup>−</sup>. = 0.30.Kf = 7.59 × 107 for BaY2−.

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