In: Chemistry
Exercise 15.82
For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH.
Part A
8.74×10−3 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
1.12×10−2 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
2.2×10−4 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
4.8×10−4 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 =
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