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Chapter 12: Q8P-b (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

(b) For 20 mL the value ofpMn2+is 2.175.

Step by step solution

01

Introduction

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

Titration  Reaction: Mn2++EDTA⇌MnY2−Kf=1013.89At   pH   8  αY4−=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+

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

=Fraction remaining × Initial concentration × Dilution factor

If 20 mL solution is added then the reaction will be 20/50 completed as the equivalence point is at 50 mL. Then the metal (Mn) concentration will be

Mn2+=50−20500.02 M2525+20=6.67×10−3 M

Therefore, the value of pMn2+

pMn2+=−logMn2+=−log6.67×10−3=2.175

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Most popular questions from this chapter

Find αZn2+if free, unprotonated [NH3] = 0.02 M.

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

ϕ=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".

A 50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.050 0 M Zn2+. What was the concentration of Ni2+ in the original solution?

Potassium ion in a 250.0 (±0.1) mL water sample was precipitated with sodium tetraphenylborate:

K++(C6H5)4B-→KB(C6H5)4(s)

The precipitate was filtered, washed, dissolved in an organic solvent, and treated with excess Hg (EDTA)2-:

4HgY2-+(C6H5)4B-+4H2O→H3BO3+4C6H5Hg++4HY3-+OH-

The liberated EDTA was titrated with 28.73 (±0.03) mL of 0.043 7 (±0.000 1) M Zn2+. Find [K+] (and its absolute uncertainty) in the original sample.

A 50.0-mL aliquot of solution containing 0.450 g of MgSO4 (FM 120.37) in 0.500 L required 37.6 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution?
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