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Chapter 12: Q2 TY (page 271)

Find [Ca²⁺] in 0.10 M CaY^2- at pH 8.00

Short Answer

Expert verified

The concentration of [Ca²⁺] is 2.3脳10^-5Min 0.10 M CaY^2- at pH 8.00.

Step by step solution

01

Introduction

The fraction of all free EDTA (in Y^4- format) is expressed in terms of $伪_{Y^{4 -}}$ . It can be expressed as

$伪_{Y^{4 -}} = \frac{[Y^{4 -}]}{[H_{6} Y^{2 +}] + [H_{5} Y^{+}] + [H_{4} Y] + [H_{3} Y^{-}] + [H_{2} Y^{2 -}] + [H Y^{3 -}] + [Y^{4 -}]}$

From table 12-1 it was found that for, $伪_{Y^{4 -}} = 4.2 脳 10^{- 3}$

From table 12-2 it was found that for CaY^2-, $K_{f} = 10^{10.65}$

02

Determine the conditional formation constant

$\begin{array}{l} K_{f}^{'} = 伪_{Y^{4 -}} K_{f} \\ K_{f}^{'} = 4.2 脳 10^{- 3} 脳 10^{10.65} \\ = 4.2 脳 10^{7.65} \end{array}$

03

Determine concentration of [Ca2+]

$\frac{C a^{2 +} + E D T A 鈬� C a Y^{2 -}}{I n i t i a l C o n c e n t r a t i o n 000.1} F i n a l C o n c e n t r a t i o n x x 0.1 - x$

From the above ICE table we can write

$\frac{[C a Y^{2 -}]}{[C a^{2 +}] [E D T A]} = K_{f}^{'} \frac{0.1 - x}{x^{2}} = 4.2 脳 10^{7.65} x = 2.3 脳 10^{- 5}$

Therefore, the concentration of [Ca²⁺] is 2.3脳10^-5M

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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 formation of the complexes ML and ML₂.Consider the titration of metal M (initial concentration = C_M, initial volume = V_M) with ligand L (concentration = C_L, volume added = V_L), which can form 1:1 and 2 : 1 complexes:

$\begin{array}{clc} M + L 饾啅 & ML & 尾_{1} = \frac{[ML]}{[M] [L]} \\ M + 2 L 饾啅 & {ML}_{2} & 尾_{2} = \frac{[{ML}_{2}]}{[M] [L]^{2}} \end{array}$

Let 伪M be the fraction of metal in the form M, 伪_ML be the fraction in the form ML, and $伪_{{ML}_{2}}$ be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

role="math" localid="1667801924683" $\begin{aligned} 伪_{M 1} = \frac{1}{1 + 尾_{1} [L] + 尾_{2} [L]^{2}} \\ 伪_{ML} = \frac{尾_{1} [L]}{1 + 尾_{1} [L] + 尾_{2} [L]^{2}} \\ 伪_{{ML}_{2}} = \frac{尾_{2} [L]^{2}}{1 + 尾_{1} [L] + 尾_{2} [L]^{2}} \end{aligned}$

The concentrations of ML and ML2are

$[ML] = 伪_{ML} \frac{C_{M} V_{M}}{V_{M} + V_{L}} [{ML}_{2}] = 伪_{{ML}_{2}} \frac{C_{M} V_{M}}{V_{M} + V_{L}}$

because $\frac{C_{M} V_{M}}{V_{M} + V_{L}}$ is the total concentration of all metal in the solution. The mass balance for ligand is

$[L] + [ML] + 2 [{ML}_{2}] = \frac{C_{M} V_{M}}{V_{M} + V_{L}}$

By substituting expressions for [ML] and [ML₂] into the mass balance, show that the master equation for a titration of metal by ligand is

$蠒 = \frac{C_{L} V_{L}}{V_{M} + V_{M}} = \frac{伪_{ML} + 2 伪_{{ML}_{2}} + \frac{L}{C_{M}}}{1 - \frac{L}{C_{L}}}$

Titration of M with L to form ML and ML₂. Use theequation from Problem 12-21, where M is Cu²⁺ and L is acetate.Consider adding 0.500 M acetate to 10.00 mL of 0.050 0 M Cu²⁺ atpH 7.00 (so that all ligand is present as ${CH}_{3} {CO}_{2}^{-}$ , not CH₃CO₂H).Formation constants forCu(CH₃CO₂)⁺ and Cu(CH₃CO₂)₂ aregiven in Appendix I. Construct a spreadsheet in which the inputis pL and the output is [L], V_L, [M], [ML], and [ML₂]. Prepare agraph showing concentrations of L, M, ML, and ML₂as V_L rangesfrom 0 to 3 mL

List four methods for detecting the end point of an EDTA Titration

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

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