Exercise 15.82 For each of the following strong base solutions, determine [OH],[

Question

Exercise 15.82

For each of the following strong base solutions, determine [OH],[H3O+], pH, and pOH.

Part A

8.74×103 M LiOH

Express your answer using three significant figures. Enter your answers numerically separated by commas.

[OH],[H3O+] =

Part B

Express your answer to three decimal places. Enter your answers numerically separated by commas.

pH,pOH =

Part C

1.12×102 M Ba(OH)2

Express your answer using three significant figures. Enter your answers numerically separated by commas.

[OH],[H3O+] =

Part D

Express your answer to three decimal places. Enter your answers numerically separated by commas.

pH,pOH =

Part E

2.2×104 M KOH

Express your answer using two significant figures. Enter your answers numerically separated by commas.

[OH],[H3O+] =

Part F

Express your answer to two decimal places. Enter your answers numerically separated by commas.

pH,pOH =

Part G

4.8×104 M Ca(OH)2

Express your answer using two significant figures. Enter your answers numerically separated by commas.

[OH],[H3O+] =

Part H

Express your answer to two decimal places. Enter your answers numerically separated by commas.

pH,pOH =

Explanation / Answer

Answer – We are given strong base and need to calculate [OH-] , [H3O+], pH, pOH

Part A) 8.74*10-3 M LiOH

We are given strong base, means

[LiOH] = [OH-] = 8.74*10-3 M

We know,

[H3O+] [OH-] = 1*10-14

So, [H3O+] = 1*10-14 / [OH-]

                  = 1*10-14 /8.74*10-3 M

                  = 1.14*10-12 M

Part B)

We know formula for pOH

pOH = -log [OH-]

         = - log 8.74*10-3 M

         = 2.06

So, pH = 14 –pOH

             = 14-2.06

              = 11.9

Part C) 1.12*10-2 M Ba(OH)2

We are given strong base, means

[Ba(OH)2] =2 [OH-] = 2*1.12*10-2 M

                            [OH-] =   2.24*10-2 M

We know,

[H3O+] [OH-] = 1*10-14

So, [H3O+] = 1*10-14 / [OH-]

                  = 1*10-14 /2.24*10-2 M

                  = 4.46*10-13 M

Part D)

We know formula for pOH

pOH = -log [OH-]

         = - log 2.24*10-2 M

         = 1.65

So, pH = 14 –pOH

             = 14-1.65

              = 12.35

Part E) 2.2*10-4 M KOH

We are given strong base, means

[KOH] = [OH-] = 2.2*10-4 M

We know,

[H3O+] [OH-] = 1*10-14

So, [H3O+] = 1*10-14 / [OH-]

                  = 1*10-14 /2.2*10-4 M

                  = 4.54*10-11 M

Part F)

We know formula for pOH

pOH = -log [OH-]

         = - log 2.2*10-4 M

         = 3.66

So, pH = 14 –pOH

             = 14-3.66

              = 10.3

Part G) 4.8*10-4 M Ca(OH)2

We are given strong base, means

[Ca(OH)2] =2 [OH-] = 2*4.8*10-4 M

                            [OH-] =   9.6*10-3 M

We know,

[H3O+] [OH-] = 1*10-14

So, [H3O+] = 1*10-14 / [OH-]

                  = 1*10-14 /2.24*10-3 M

                  = 1.04*10-11 M

Part H)

We know formula for pOH

pOH = -log [OH-]

         = - log 9.6*10-3 M

         = 3.018

So, pH = 14 –pOH

             = 14-3.018

              = 10.98

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