5 0.0030 0.0015 0.0005 6 0.0030 0.0057 0.0089 7 0.0086 0.0057 0.0089Show the Calculations of Qc and Kc for Exp. 5-7Please help me out. Thanks tpermehe touuneDemiy 200-Gener OemiryDept of hi Soencm&Engneeringwibur Wrght CotegShow the Cakulations of aand K. for t. 57CaletionaICon ofCaktaion ofa Wilbur Wright CollegeREPORTTable 1: Initial & Equilibrium Concentrations[Fe") (M)[SCN] (M)(FESCN) (M)(Fe"l (M)5.58 10 M(SCN J (M)(FESCN (M)4.42 *10 MExp.5.58 *10 M10.0100.0103.38 *10 M3.38 * 10 M1.62 10 M20.00500.00504.55 10 M4.55 10 M2.95 10 M30.00500.00500.00254.99* 10 M8.49 10 M6.01 10 M40.00250.00600.00852.9* 20 M1.41 10 M5.85 10 M50.00300.00150.000560.00300.00570.00895.50* 10 M8.20* 10 M6.40* 10 M9.22* 10 M6.32* 10 M8.28* 10 M0.00860.00570.0089

The equilibrium reaction between the Fe3+ ion and SCN- ion to given an iron thiocyanate complex can…

Answered: 5 0.0030 0.0015 0.0005…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter20: Applications Of Oxidation/reduction Titrations

Section: Chapter Questions

Problem 20.17QAP

See similar textbooks

Related questions

Concept explainers

Question

100%

0.0030

0.0015

0.0005

0.0030

0.0057

0.0089

0.0086

0.0057

0.0089

Show the Calculations of Q_cand K_cfor Exp. 5-7

Please help me out. Thanks

tpermehe touune
Demiy 200-Gener Oemiry
Dept of hi Soencm&Engneering
wibur Wrght Coteg
Show the Cakulations of aand K. for t. 57
Caletiona
ICon of
Caktaion ofa

Wilbur Wright College
REPORT
Table 1: Initial & Equilibrium Concentrations
[Fe") (M)
[SCN] (M)
(FESCN) (M)
(Fe"l (M)
5.58 10 M
(SCN J (M)
(FESCN (M)
4.42 *10 M
Exp.
5.58 *10 M
1
0.010
0.010
3.38 *10 M
3.38 * 10 M
1.62 10 M
2
0.0050
0.0050
4.55 10 M
4.55 10 M
2.95 10 M
3
0.0050
0.0050
0.0025
4.99* 10 M
8.49 10 M
6.01 10 M
4
0.0025
0.0060
0.0085
2.9* 20 M
1.41 10 M
5.85 10 M
5
0.0030
0.0015
0.0005
6
0.0030
0.0057
0.0089
5.50* 10 M
8.20* 10 M
6.40* 10 M
9.22* 10 M
6.32* 10 M
8.28* 10 M
0.0086
0.0057
0.0089

Expert Solution

Step 1

The equilibrium reaction between the Fe³⁺ ion and SCN^- ion to given an iron thiocyanate complex can be written as follows.

${Fe}^{2 +} (aq) + {SCN}^{-} ⇌ {FeSCN}^{-} (aq)$

Q_c is the reaction quotient for the given reaction, which can be defined as the ratio of the molar concentration of product to that of the molar concentration of the reactant at that instant of time raised that are raised to their respective stoichiometric coefficient.

So, for this the reaction quotient Q_c can be written as the following expression.

$Q_{c} = \frac{[{FeSCN}^{-}]}{[{Fe}^{2 +}] [{SCN}^{-}]}$

The equilibrium constant K_c is similar to the reaction quotient but differs in terms of the concentrations that are used to find the ratio. In equilibrium constant, the equilibrium concentration of the products and reactants are used. For equilibrium concentration, subscript of 'eq' can be used to differentiate the concentration temp used for the expression of equilibrium constant.

$K_{c} = \frac{{[{FeSCN}^{-}]}_{eq}}{{[{Fe}^{2 +}]}_{eq} \cdot {[{SCN}^{-}]}_{eq}}$

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

5 0.0030 0.0015 0.0005 6 0.0030 0.0057 0.0089 7 0.0086 0.0057 0.0089 Show the Calculations of Qc and Kc for Exp. 5-7