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Chapter 12: Q8P-e (page 283)

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

(e) For 49.9 mL the value ofpMn2+is 4.87.

Step by step solution

01

Introduction

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

Titration鈥夆赌�Reaction:鈥�Mn2++EDTA鈬�MnY2鈭�Kf=1013.89At鈥夆赌夆��pH鈥夆赌夆��8鈥夆赌�伪Y4鈭�=4.2脳10鈭�3鈥�Table鈥�12鈭�1

02

Determine equilibrium constant

Kf'=伪Y4鈭�脳Kf=4.2脳10鈭�3脳1013.89=3.3脳1011

03

Determine the value of pMn2+

The concentration of the remaining productcan be calculated using the following equation

=Fraction remaining 脳 Initial concentration 脳 Dilution factor

If 49 .9mL solution is added then the reaction will be 49.9/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Mn2+=50鈭�49.9500.02鈥�M2525+49.9=1.34脳10鈭�5鈥�M

Therefore, the value of pMn2+

pMn2+=鈭�logMn2+=鈭�log1.34脳10鈭�5=4.87

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Most popular questions from this chapter

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

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

At what pH does 伪y4-=0.50?

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

蠒=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".

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL
See all solutions

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