Chapter 15, Problem 112CP

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

Chapter
Section

### Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

# Consider a solution formed by mixing 50.0 mL of 0.100 M H2SO4, 30.0 mL of 0.100 M HOCI, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ba(OH)2, and 10.0 mL of 0.150 M KOH. Calculate the pH of this solution.

Interpretation Introduction

Interpretation: A solution consisting of different bases and acids is given. The value of pH of the given solution is to be calculated.

Concept introduction: The strong bases have tendency to undergo complete dissociation while the weak bases do not undergo complete dissociation. Similar is the case with the strong and weak acid. The dissociation of acid and base takes place in a step wise manner.

Explanation

Explanation

Given

The concentration of NaOH is 0.200M .

The concentration of KOH is 0.150M .

The concentration of Ba(OH)2 is 0.100M .

The concentration of H2SO4 is 0.100M .

The concentration of HOCl is 0.100M .

The volume of NaOH is 25.0mL .

The volume of KOH is 10.0mL .

The volume of Ba(OH)2 is 25.0mL .

The volume of H2SO4 is 50.0mL .

The volume of HOCl is 30mL .

The conversion of mL into L is done as,

1mL=0.001L

The conversion of all the given species into l is done as,

50.0mL=50.0×0.001L=0.050L25.0mL=25.0×0.001L=0.025L

Similarly the other volumes are converted as,

30.0mL=30.0×0.001L=0.030L10.0mL=10.0×0.001L=0.010L

All the given bases are strong; therefore they will get completely dissociated. The dissociation of the bases is shown as,

NaOHNa++OHBa(OH)2Ba2++2OHKOHK++OH

Now on dissociation the concentration of Na+ and OH produced is equal to the concentration of NaOH and similarly in case of KOH the concentration of K+ and OH is equal to the concentration of KOH . But in case of Ba(OH)2 , the concentration of Ba2+ is same as that of Ba(OH)2 but the concentration of OH is twice the concentration of Ba(OH)2 .

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (1)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres (2)

Substitute the value of concentration and of OH obtained from each species and the value of volume to obtain the overall concentration of OH .

The number of moles of OH obtained from NaOH are given as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.025L=0.005moles

The number of moles of OH obtained from KOH are given as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.150M×0.010L=0.0015moles

The number of moles of OH obtained from Ba(OH)2 are given as,

Numberofmoles=Concentration×Volumeofsolutioninlitres=0.200M×0.025L=0.005moles

Total number of moles of OH =(0.005+0.0015+0.005)moles=0.0115moles

Total volume of the solution =(0.025+0.025+0.010)L=0.06L

Substitute the value of number of moles of OH and total volume in the equation (1).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.0115moles0.06L=0.1916M

One of the given acid H2SO4 is a strong acid and being diprotic it undergoes dissociation two times.

The product from first dissociation of H2SO4 is shown as,

H2SO4H++HSO4

Now, the produced HSO4 acts as a weak acid whose dissociation is shown as,

HSO4H++SO42

Make the ICE table for the dissociation reaction of HSO4 .

HSO4H+(aq)SO42(aq)Initial(M):0.10000Change(M):xxxEquilibrium(M):0.100xxx

The equilibrium ratio for the given reaction is,

Ka=[SO42][H+][HSO4]

Substitute the calculated concentration values in the above expression.

Ka=[SO42][H+][HSO4]1.2×102=(x)(x)(0.100x)M1.2×102×(0.100x)=(x)(x)0.00120.012x=x2

The equation is further simplified as,

x2+0.012x0.0012=0

The quadratic equation is solved as,

d=b24ac

Where,

• b is the coefficient of x .
• a is the coefficient of x2 .
• c is the coefficient of x0 .
• d is the discriminant of the equation.

Substitute the value of a , b and c in the above equation as,

d=b24ac=(0

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