Chemistry Smartwork Access Code Fourth Edition
Chemistry Smartwork Access Code Fourth Edition
4th Edition
ISBN: 9780393521368
Author: Gilbert
Publisher: NORTON
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Chapter 17, Problem 17.52QP
Interpretation Introduction

Interpretation: The value of pH for 1.00M Cu(NO3)2 is to be calculated.

Concept introduction: The potential of H+ ions is called as pH. The value of pH is less than 7 for acidic solutions and greater than 7 for basic solutions. For neutral solution the value of pH is 7.

To determine: The value of pH for 1.00M Cu(NO3)2.

Expert Solution & Answer
Check Mark

Answer to Problem 17.52QP

The value of pH for 1.00M Cu(NO3)2 is 3.76_.

Explanation of Solution

The concentration of Cu(NO3)2 is 1.00M.

The chemical equation for the dissociation of Cu(NO3)2 is,

Cu(NO3)2(aq)Cu2+(aq)+2NO3(aq)

In general Cu2+(aq) ions is in hydrated form that is [Cu(H2O)6]2+. Dissociation of this ion in water is,

[Cu(H2O)6]2+(aq)+H2O(l)H3O+(aq)+[Cu(H2O)5(OH)]+(aq)

The Ka of Cu(NO3)2 is 3×108.

The ICE table for the dissociation of Cu(NO3)2 is,

[Cu(H2O)6]2+(aq)[Cu(H2O)5(OH)]+(aq)+H3O+(aq)I1.0000CxxxE1.00xxx

The concentrations of both [Cu(H2O)5(OH)]+ and H3O+ are equal and assumed to be x and is calculated by the equilibrium constant expression,

Ka=[[Cu(H2O)5(OH)]+][H3O+][[Cu(H2O)6]2+]

Substitute the values of dissociation constant and concentration of base in the above expression to calculate the concentration of H+ ions.

3×108=x21.00xx2=3×108(1.00x)x2=3×1083×108x0=x2+3×108x3×108

Simplify the above equation by the formula,

x=b±b24ac2a

Substitute the values of a and b in the above formula to calculate the value of x.

x=(3×108)±(3×108)24×1×(3×108)2×1=3×108±9×1016+1.2×1072=3×108+3.464×1042=1.73×104M

Hence, the concentration of H3O+ is 1.73×104M.

The pH of a solution is calculated by the formula,

pH=log[H3O+]

Substitute the value of [H3O+] in the above formula to calculate the value of pH.

pH=log(1.73×104)=3.76_

Hence, the pH of a solution is 3.76_.

Conclusion

The pH of a Cu(NO3)2 solution is 3.76_.

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Chapter 17 Solutions

Chemistry Smartwork Access Code Fourth Edition

Ch. 17 - Prob. 17.5VPCh. 17 - Prob. 17.6VPCh. 17 - Prob. 17.7VPCh. 17 - Prob. 17.8VPCh. 17 - Prob. 17.9VPCh. 17 - Prob. 17.10VPCh. 17 - Prob. 17.11QPCh. 17 - Prob. 17.12QPCh. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - Prob. 17.15QPCh. 17 - Prob. 17.16QPCh. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - Prob. 17.20QPCh. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - Prob. 17.23QPCh. 17 - Prob. 17.24QPCh. 17 - Prob. 17.25QPCh. 17 - Prob. 17.26QPCh. 17 - Prob. 17.28QPCh. 17 - Prob. 17.29QPCh. 17 - Prob. 17.30QPCh. 17 - Prob. 17.31QPCh. 17 - Prob. 17.32QPCh. 17 - Prob. 17.33QPCh. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - Prob. 17.36QPCh. 17 - Prob. 17.37QPCh. 17 - Prob. 17.38QPCh. 17 - Prob. 17.39QPCh. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - Prob. 17.42QPCh. 17 - Prob. 17.43QPCh. 17 - Prob. 17.44QPCh. 17 - Prob. 17.45QPCh. 17 - Prob. 17.46QPCh. 17 - Prob. 17.47QPCh. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - Prob. 17.52QPCh. 17 - Prob. 17.53QPCh. 17 - Prob. 17.54QPCh. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - Prob. 17.57QPCh. 17 - Prob. 17.58QPCh. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - Prob. 17.64QPCh. 17 - Prob. 17.65QPCh. 17 - Prob. 17.66QPCh. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - Prob. 17.70QPCh. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - Prob. 17.73QPCh. 17 - Prob. 17.74QPCh. 17 - Prob. 17.75QPCh. 17 - Prob. 17.76QPCh. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - Prob. 17.79QPCh. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - Prob. 17.84QPCh. 17 - Prob. 17.85APCh. 17 - Prob. 17.86APCh. 17 - Prob. 17.87APCh. 17 - Prob. 17.88APCh. 17 - Prob. 17.89APCh. 17 - Prob. 17.90APCh. 17 - Prob. 17.91AP
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