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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

Short Answer

Expert verified

(h) For 55 mL the value of pMn2+is 10.52.

Step by step solution

01

Introduction

Equations and data obtained in order to proceed for calculation are as follows

Titration â¶Ä‰Reaction: Mn2++EDTA⇌MnY2−Kf=1013.89At â¶Ä‰â€‰pH â¶Ä‰â€‰8 â¶Ä‰Î±Y4−=4.2×10−3 Table 12−1

02

Determine equilibrium constant

Kf'=αY4−×Kf=4.2×10−3×1013.89=3.3×1011

Equivalence point=50 mL

03

Determine the value of pMn2+

Past equivalence point (55mL) we need to calculate the metal concentration

There is 5 mL of Excess EDTA

 EDTA= 525+55×0.01M=6.25×10−4MMnY2−= 2525+55×0.02M=6.25×10−3M

Kf'=MnY2−Mn2+EDTA3.3×1011=6.25×10−3Mn2+6.25×10−4Mn2+=3.03×10−11M

Therefore, the value of pCu2+

pMn2+=−logMn2+=−log3.03×10−13=10.52

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Most popular questions from this chapter

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

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

Ï•=CETDAVETDACMVM=1+Kf"[M]free-[M]free+Kf"[M]freeCMKf"[M]free+[M]free+Kf"[M]free2CETDA

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 Kf".

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

If back titration required 13.00 mL Zn2+, what was the original concentration of Ni2+?

Sulfide ion was determined by indirect titration with EDTA. To a solution containing 25.00 mL of 0.04332 M Cu(ClO4)2 plus 15 mL of 1 M acetate buffer (pH 4.5) were added 25.00 mL of unknown sulfide solution with vigorous stirring. The CuS precipitate was filtered and washed with hot water. Ammonia was added to the filtrate (which contained excess Cu2+) until the blue color of Cu(NH3)42+ was observed. Titration of the filtrate with 0.039 27 M EDTA required 12.11 mL to reach the murexide end point. Calculate the molarity of sulfide in the unknown.

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