Exploring Chemical Analysis
Exploring Chemical Analysis
5th Edition
ISBN: 9781429275033
Author: Daniel C. Harris
Publisher: Macmillan Higher Education
Question
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Chapter 11, Problem 11.14P

(a)

Interpretation Introduction

Interpretation:

Concentration of each species in below diagram has to be calculated.

Exploring Chemical Analysis, Chapter 11, Problem 11.14P , additional homework tip  1

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged for of equilibrium constant, Ka. Consider an equation of dissociation of an acid as follows:

  HAH++A

Formula to calculate pH is as follows:

  pH=pKa+log([A][HA])

Here,

pKa is acid dissociation constant of weak acid HA.

[A] is concentration of conjugate base A

[HA] is concentration of acid HA.

(a)

Expert Solution
Check Mark

Explanation of Solution

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK1+log[HA][H2A]        (1)

Substitute pK1 for pH in equation (1).

  pK1=pK1+log[HA][H2A]log[HA][H2A]=0[HA][H2A]=1

At pH equal to pK1, concentration of HA is equal to H2A.

In accordance to Henderson Hasselbach equation pH1 is calculated as follows:

  pH1=pK1+log[HA][H2A]        (1)

In accordance to Henderson Hasselbach equation pH2 is calculated as follows:

  pH2=pK2+log[A2][HA]        (2)

The expression for final pH is calculated as follows:

  pH=12(pH1+pH2)        (3)

Substitute pK1+log[HA][H2A] for pH1 and pK2+log[A2][HA] for pH2 in equation (3).

  pH=12(pK1+log[HA][H2A]+pK2+log[A2][HA])=12(pK1+pK2)+12(log[HA][H2A]+log[A2][HA])=12(pK1+pK2)+12log([A2][H2A])

Substitute pH for 12(pK1+pK2) in above equation.

  pH=pH+12log([A2][H2A])log([A2][H2A])=0[A2][H2A]=1

At pH equal to 12(pK1+pK2), concentration of A2 is equal to H2A.

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK2+log[A2][HA]        (2)

Substitute pK2 for pH in equation (2).

  pK2=pK2+log[A2][HA]log[A2][HA]=0[A2][HA]=1

At pH equal to pK2, concentration of A2 is equal to HA.

(b)

Interpretation Introduction

Interpretation:

Analogous diagrams for monoprotic and triprotic systems have to be drawn.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

In diagram of monoprotic system if pH is equal to pKa then solution is considered as neutral. If value of pH is less than pKa, solution becomes acidic and value of pH is more than pKa, solution becomes basic. Hence, diagram is as follows:

Exploring Chemical Analysis, Chapter 11, Problem 11.14P , additional homework tip  2

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK1+log[A][HA]        (4)

Substitute pK1 for pH in equation (4).

  pK1=pK1+log[A][HA]log[A][HA]=0[A][HA]=1

At pH equal to pK1, concentration of A is equal to HA.

In diagram of triprotic system diagram consist of pK1, pK2 and pK3 that is related to pH in acidic, neutral and basic region. Hence, diagram is as follows:

Exploring Chemical Analysis, Chapter 11, Problem 11.14P , additional homework tip  3

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK1+log[H2A][H3A]        (5)

Substitute pK1 for pH in equation (5).

  pK1=pK1+log[H2A][H3A]log[H2A][H3A]=0[H2A][H3A]=1

At pH equal to pK1, concentration of H2A is equal to H3A.

In accordance to Henderson Hasselbach equation pH1 is calculated as follows:

  pH1=pK1+log[H2A][H3A]        (6)

In accordance to Henderson Hasselbach equation pH2 is calculated as follows:

  pH2=pK2+log[HA2][H2A]        (7)

The expression for final pH is calculated as follows:

  pH=12(pH1+pH2)        (8)

Substitute pK1+log[H2A][H3A] for pH1 and pK2+log[HA2][H2A] for pH2 in equation (8).

  pH=12(pK1+log[H2A][H3A]+pK2+log[HA2][H2A])=12(pK1+pK2)+12(log[H2A][H3A]+log[HA2][H2A])=12(pK1+pK2)+12log([HA2][H3A])

Substitute pH for 12(pK1+pK2) in above equation.

  pH=pH+12log([HA2][H3A])log([HA2][H3A])=0[HA2][H3A]=1[HA2]=[H3A]

At pH equal to 12(pK1+pK2), concentration of HA2 is equal to H3A.

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK2+log[HA2][H2A]        (9)

Substitute pK2 for pH in equation (9).

  pK2=pK2+log[HA2][H2A]log[HA2][H2A]=0[HA2][H2A]=1[HA2]=[H2A]

At pH equal to pK2, concentration of HA2 is equal to H2A.

In accordance to Henderson Hasselbach equation pH2 is calculated as follows:

  pH2=pK2+log[HA2][H2A]        (10)

In accordance to Henderson Hasselbach equation pH3 is calculated as follows:

  pH3=pK3+log[A3][HA2]        (11)

The expression for final pH is calculated as follows:

  pH=12(pH2+pH3)        (12)

Substitute pK2+log[HA2][H3A] for pH2 and pK3+log[A3][HA2] for pH3 in equation (12).

  pH=12(pK2+log[HA2][H3A]+pK3+log[A3][HA2])=12(pK2+pK3)+12(log[HA2][H2A]+log[A3][HA2])=12(pK2+pK3)+12log([A3][H2A])

Substitute pH for 12(pK2+pK3) in above equation.

  =pH+12log([A3][H2A])log([A3][H2A])=0[A3][H2A]=1[A3]=[H2A]

At pH equal to 12(pK2+pK3), concentration of A3 is equal to H2A.

In accordance to Henderson Hasselbach equation pH is calculated as follows:

  pH=pK3+log[A3][HA2]        (13)

Substitute pK3 for pH in equation (13).

  pK3=pK3+log[A3][HA2]log[A3][HA2]=0[A3][HA2]=1[A3]=[HA2]

At pH equal to pK3, concentration of A3 is equal to HA2.

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