Q8P-i Consider the titration of 25.0 m... [FREE SOLUTION]

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Chapter 12: Q8P-i (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

(i) For 60 mL the value of pMn²⁺is 10.82.

Step by step solution

Introduction

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

$\begin{matrix} T i t r a t i o n 鈥夆赌� Re a c t i o n : 鈥� M n^{2 +} + E D T A 鈬� M n Y^{2 鈭�} \\ K_{f} = 10^{13.89} \\ A t 鈥夆赌夆€� p H 鈥夆赌夆€� 8 鈥夆赌� 伪_{Y^{4 鈭�}} = 4.2 脳 10^{鈭� 3} 鈥� (T a b l e 鈥� 12 鈭� 1) \end{matrix}$

Determine equilibrium constant

$\begin{matrix} K_{f}^{'} = 伪_{Y^{4 鈭�}} 脳 K_{f} \\ = 4.2 脳 10^{鈭� 3} 脳 10^{13.89} \\ = 3.3 脳 10^{11} \end{matrix}$

Equivalence point=50 mL

Determine the value of pMn2+

Past equivalence point (60mL) we need to calculate the metal concentration

There is 10 mL of Excess EDTA

$\begin{matrix} E D T A = 鈥� (\frac{10}{25 + 60} 脳 0.01 M) \\ = 1.176 脳 10^{鈭� 3} M \\ M n Y^{2 鈭�} = 鈥� (\frac{25}{25 + 60} 脳 0.02 M) \\ = 5.88 脳 10^{鈭� 3} M \end{matrix}$

$\begin{matrix} K_{f}^{'} = \frac{[M n Y^{2 鈭�}]}{[M n^{2 +}] [E D T A]} \\ 3.3 脳 10^{11} = \frac{[5.88 脳 10^{鈭� 3}]}{[M n^{2 +}] [1.176 脳 10^{鈭� 3}]} \\ M n^{2 +} = 1.5 脳 10^{鈭� 11} M \end{matrix}$

Therefore, the value of pCu²⁺

^{$\begin{matrix} p M n^{2 +} = 鈭� \log (M n^{2 +}) \\ = 鈭� \log (1.5 脳 10^{鈭� 11}) \\ = 10.82 \end{matrix}$}

Graph

The titration curve combining all the data from SID 135385-12-8P-a to SID 135385-12-8P-i was obtained

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91影视

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+

Graph

One App. One Place for Learning.

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91影视

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+

Graph

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Organic Chemistry

Nuclear Chemistry

Chemistry Branches

Physical Chemistry

Kinetics

Study anywhere. Anytime. Across all devices.

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