EBK CHEMISTRY: THE MOLECULAR NATURE OF
EBK CHEMISTRY: THE MOLECULAR NATURE OF
8th Edition
ISBN: 9781259915505
Author: SILBERBERG
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 18, Problem 18.81P
Interpretation Introduction

Interpretation:

The [H2C3H2O4],  [HC3H2O4],  [C3H2O42], [H3O+], pH, [OH-], and pOH have to be calculated in 0.20 M solution of the diprotic malonic acid.

Concept introduction:

An equilibrium constant (K) is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  HA(aq)+H2O(l)H3O+(aq)+A(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  Ka=[H3O+][A][HA]                                                                                                   (1)

An equilibrium constant (K) with subscript a indicate that it is an equilibrium constant of an acid in water.

Expert Solution & Answer
Check Mark

Explanation of Solution

Given,

Solution of 0.2Mmalonic acid (H2C3H2O4).

Malonic acid is diprotic acid therefore it can donate the two protons to the water.

The balance equation is given below,

  H2C3H2O4 (aq)+H2O(l)H3O(aq) +   HC3H2O4(aq)The value Ka1is calculating by using following formula, Ka1 =[HC3H2O4][H3O][H2C3H2O4]given,Ka1 =[HC3H2O4][H3O][H2C3H2O4]= 1.4×103

The formed HC3H2O4 ion can donate another proton, therefore the balance equation is given below,

HC3H2O4 (aq)+H2O(l)H3O(aq) +   C3H2O42(aq)The value Ka2is calculating by using following formula, Ka2 =[C3H2O42][H3O][HC3H2O4]given,Ka2 =[C3H2O42][H3O][HC3H2O4]= 2.0×106

From the dissociation, Ka1Ka2, therefore the first dissociation forms complete H3O ion.

 0.2Mmalonic acid (H2C3H2O4) Solution.

Therefore,

ICE table:

  H2C3H2O4 (aq)+H2O(l)H3O(aq) +   HC3H2O4(aq)

Initial concentration0.2 M-00
Change -x + x+ x
At equilibrium0.2-x xx

The initial concentration is 0.2Mmalonic acid (H2C3H2O4).

  H2C3H2O4 (aq)+H2O(l)H3O(aq) +   HC3H2O4(aq)The value Ka1is calculating by using following formula, Ka1 =[HC3H2O4][H3O][H2C3H2O4]given,Ka1 =[HC3H2O4][H3O][H2C3H2O4]= 1.4×103

Let consider,

  Ka1 =[HC3H2O4][H3O][H2C3H2O4]= 1.4×103x2(0.2x)= 1.4×103x2=2.8×104x=0.0167

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100Percent dissociated =  0.01670.2×100Percent dissociated =  8.3%

Percentage of error is high and it is not valid, the dissociation of H2C3H2O4 is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression can be solved using the quadratic formula.

The quadratic formula is given below,

  x2+1.4×103x2.8×104=0a=1,b=1.4×103,c=2.8×104x=b±b24ac2ax=(1.4×103)±(1.4×103)24(1)(2.8×104)2(1)x=1.6×102 M=  [H3O+],[HC3H2O4]=[H3O+]=1.6×102 M

Hence, [HC3H2O4-]=[H3O+]=1.6×10-2 M

Therefore,

  pH=-log[H3O+]pH=-log(1.6×102 M)pH=1.81

 pH of 0.2 M H2C3H2O4 is 1.81..

The [OH-] value is calculating by using following formula,

  [OH-] =  Kw[H3O+][OH-] =  1×10-141.6×102[OH-] =  6.25×10-13

Hence, [OH-] is 6.25×10-13.

Therefore,

  pOH=-log[OH]pOH=-log(6.25×10-13)pOH=12.20

pOH of the solution is 12.20.

The Malonic acid concentration at equilibrium is [H2C3H2O4]init – [H2C3H2O4]dissoc

The Malonic acid concentration at equilibrium is 0.2 – 1.6×102M

The Malonic acid concentration at equilibrium is 0.184 M.

Therefore, the Ka2 expression is given below for calculating [C3H2O42], and [HC3H2O4] and [H3O+] come mostly from the first dissociation. This new calculation will have a new x value,

Therefore,

ICE table:

  HC3H2O4 (aq)+H2O(l)H3O(aq) +   C3H2O42(aq)

Initial concentration1.6×102 M-1.6×102 M0
Change -x + x+ x
At equilibrium1.6×102 M -x 1.6×102 M+xx

  The value Ka2is calculating by using following formula, Ka2 =[C3H2O42][H3O][HC3H2O4]= 2.0×106Ka2=(1.6×102+x)×x (1.6×102x)=2.0×106Ka2=(1.6×102)×x (1.6×102x)=2.0×105[C3H2O42]2.0×105M

X value is very small, hence not necessary to calculate the HC3H2O4 andH3O.

Hence, [C3H2O42] is 2.0×10-5M.

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Chapter 18 Solutions

EBK CHEMISTRY: THE MOLECULAR NATURE OF

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