General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 20, Problem 20.50P
Interpretation Introduction

Interpretation:

The equilibrium concentration values of OH-(aq), HNO2(aq), NO2-(aq) and H3O+(aq) has to be calculated. Also the pH of solution should be determined at 25oC.

Concept introduction:

Ionic-product constant for water: It is the hydronium ion concentration times the OH concentration present in the solution.

H2O(l)+H2O(l)H3O+(aq)+OH(aq)K=[ H3O+ ][ OH ][ H2O ][ H2O ][ H3O+ ]=[ OH ]=1.00×107MKW=K[ H2O ][ H2O ]=[ H3O+ ][ OH ]KW=1.00×1014MKW=Ka.Kb

Expert Solution & Answer
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Answer to Problem 20.50P

The concentration value of OH-(aq) is 2.1×106M.

The concentration value of HNO2(aq) is 2.1×106M.

The concentration value of NO2-(aq) is 0.25M.

The concentration value of H3O+(aq) is 4.79×10-9 M.

The pH value is 8.38.

Explanation of Solution

Given: 0.25 M in NaNO2(aq).

The reaction between NaNO2(aq) and water is as follows,

    NaNO2(s) + H2O(l)HNO2(aq) + OH-(aq)

The Ka value for HNO2(aq) is 5.6×10-4 M. Now, the Kb value is determined by using equation, Kw= Ka×Kb.

    Kw= Ka×KbKb=KwKa= 1.00×1014 M5.6×10-4 M= 1.79×1011

The ICE table for reaction of NO2- with water is shown as below,

     NO2-+ H2HNO2 + OH-Initial 0.25M         0           0Change              -x                          +x        +xEquilibrium     0.25-x                       x           x

Using calculated Kb value the concentration of HNO2 and OH- are determined.

    Kb=[HNO2][OH-][NO2-]1.79×1011= x × x0.25 - x1.79×1011= 20.25=2.1×106

    [HNO2] = [OH-]= 2.1×106M[NO2-] = 0.25 - x = 0.25 - 2.1×106[NO2-]= 0.25M

Now, the pH value is calculated by using above determined hydroxide ions concentration.

    pOH = -log [2.1×106]= 5.68pH = 14 - pOH= 14 - 5.68= 8.32

Using above calculated pH the concentration of [H3O+] ions are calculated.

    pH = -log [H3O+][H3O+]=108.32[H3O+] = 4.79×10-9 M

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Chapter 20 Solutions

General Chemistry

Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Prob. 20.25PCh. 20 - Prob. 20.26PCh. 20 - Prob. 20.27PCh. 20 - Prob. 20.28PCh. 20 - Prob. 20.29PCh. 20 - Prob. 20.30PCh. 20 - Prob. 20.31PCh. 20 - Prob. 20.32PCh. 20 - Prob. 20.33PCh. 20 - Prob. 20.34PCh. 20 - Prob. 20.35PCh. 20 - Prob. 20.36PCh. 20 - Prob. 20.37PCh. 20 - Prob. 20.38PCh. 20 - Prob. 20.39PCh. 20 - Prob. 20.40PCh. 20 - Prob. 20.41PCh. 20 - Prob. 20.42PCh. 20 - Prob. 20.43PCh. 20 - Prob. 20.44PCh. 20 - Prob. 20.45PCh. 20 - Prob. 20.46PCh. 20 - Prob. 20.47PCh. 20 - Prob. 20.48PCh. 20 - Prob. 20.49PCh. 20 - Prob. 20.50PCh. 20 - Prob. 20.51PCh. 20 - Prob. 20.52PCh. 20 - Prob. 20.53PCh. 20 - Prob. 20.54PCh. 20 - Prob. 20.55PCh. 20 - Prob. 20.56PCh. 20 - Prob. 20.57PCh. 20 - Prob. 20.59PCh. 20 - Prob. 20.60PCh. 20 - Prob. 20.61PCh. 20 - Prob. 20.62PCh. 20 - Prob. 20.63PCh. 20 - Prob. 20.64PCh. 20 - Prob. 20.65PCh. 20 - Prob. 20.66PCh. 20 - Prob. 20.67PCh. 20 - Prob. 20.68PCh. 20 - Prob. 20.69PCh. 20 - Prob. 20.70PCh. 20 - Prob. 20.71PCh. 20 - Prob. 20.72PCh. 20 - Prob. 20.73PCh. 20 - Prob. 20.74PCh. 20 - Prob. 20.75PCh. 20 - Prob. 20.76PCh. 20 - Prob. 20.77PCh. 20 - Prob. 20.78PCh. 20 - Prob. 20.79PCh. 20 - Prob. 20.80PCh. 20 - Prob. 20.81PCh. 20 - Prob. 20.82PCh. 20 - Prob. 20.83PCh. 20 - Prob. 20.84PCh. 20 - Prob. 20.85PCh. 20 - Prob. 20.86PCh. 20 - Prob. 20.87PCh. 20 - Prob. 20.88PCh. 20 - Prob. 20.89PCh. 20 - Prob. 20.90PCh. 20 - Prob. 20.91PCh. 20 - Prob. 20.92PCh. 20 - Prob. 20.94P
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