Chapter 12: Q8P-d (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

(d) For 49 mL the value ofpMn²⁺is 3.87.

Step by step solution

Introduction

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

$\begin{matrix} T i t r a t i o n Re a c t i o n : M n^{2 +} + E D T A ⇌ M n Y^{2 -} \\ K_{f} = 10^{13.89} \\ A t p H 8 α_{Y^{4 -}} = 4.2 \times 10^{- 3} (T a b l e 12 - 1) \end{matrix}$

Determine equilibrium constant

$\begin{matrix} K_{f}^{'} = α_{Y^{4 -}} \times K_{f} \\ = 4.2 \times 10^{- 3} \times 10^{13.89} \\ = 3.3 \times 10^{11} \end{matrix}$

Equivalence point=50 mL

Determine the value of pMn2+

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

=Fraction remaining × Initial concentration × Dilution factor

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

$\begin{matrix} [M n^{2 +}] = (\frac{50 - 49}{50}) (0.02 M) (\frac{25}{25 + 49}) \\ = 1.35 \times 10^{- 4} M \end{matrix}$

Therefore, the value of pMn²⁺

^{$\begin{matrix} p M n^{2 +} = - \log (M n^{2 +}) \\ = - \log (1.35 \times 10^{- 4}) \\ = 3.87 \end{matrix}$}

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Introduction

Determine equilibrium constant

Determine the value of pMn2+

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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

Step by step solution

Introduction

Determine equilibrium constant

Determine the value of pMn2+

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Ionic and Molecular Compounds

Physical Chemistry

Chemical Reactions

Chemistry Branches

Inorganic Chemistry

Study anywhere. Anytime. Across all devices.