Question

For each strong base solution, determine [H3O+], [OH−], pH, and pOH.

A)8.89×10⁻³ M LiOH

[OH−],[H3O+] =

b)8.89×10⁻³ M LiOH

pH,pOH=

C)1.15×10⁻² M Ba(OH)2

[OH−],[H3O+]=

D)1.15×10⁻² M Ba(OH)2

pH,pOH =

E)1.8×10⁻⁴ M KOH

[OH−],[H3O+]=
F)1.8×10⁻⁴ M KOH

pH,pOH =

G)4.9×10⁻⁴ M Ca(OH)2

[OH−],[H3O+] =

H)4.9×10⁻⁴ M Ca(OH)2

pH,pOH =

Answer 1

Answer A:

LiOH is a strong base. Thus, [OH ^–] = 8.89 x 10 ^{– 3} M

[H₃O⁺] = K_w/[OH ^–] = 1.12 x 10 ^{– 12} M

Answer B:

pOH = - log [OH ^– ] = - log 8.89 x 10 ^{– 3} = 2.05

pH = 14 – pOH = 11.95

Answer C:

Ba(OH)₂ is a strong base and it is di-acidic base.

i.e., each one mole of base molecule produces 2 moles OH ^–.

Thus, [OH ^–] =2 x 1.15 x 10 ^{– 2} M = 2.3 x 10 ^{– 2} M

[H₃O⁺] = K_w/[OH ^–] = 4.34 x 10 ^{– 13} M

Answer D:

pOH = - log [OH ^– ] = - log 2.3 x 10 ^{– 2} = 1.64

pH = 14 – pOH = 12.36

Answer E:

KOH is a strong base. Thus, [OH ^–] = 1.8 x 10 ^{– 4} M

[H₃O⁺] = K_w/[OH ^–] = 5.55 x 10 ^{– 11} M

Answer F:

pOH = - log [OH ^– ] = - log 1.8 x 10 ^{– 4} = 3.74

pH = 14 – pOH = 10.26

Answer G:

Ca(OH)₂ is not a strong base. First dissociation is 100 % but the second dissociation is weak and has pKa=2.37.

Thus, [OH ^–] = 4.9 x 10 ^{– 4} + 4.4 x 10 ^{– 4} = 9.3 x 10 ^{– 4} M

[H₃O⁺] = K_w/[OH ^–] = 1.07 x 10 ^{– 11} M

Answer H:

pOH = - log [OH ^– ] = - log 9.3 x 10 ^{– 4} = 3.03

pH = 14 – pOH = 10.97