Answer to Question #181869 in General Chemistry for Roberto

Question

Using data from Appendix D in your textbook, calculate [OH⁻] and pH for each of the following solutions

1. 0.10 M NaBrO
2. 0.080 M NaHS
3. A mixture that is 0.045 M in CH3COONa and 0.055 M in (CH3COO)2Ba

Answer 1

1. 0.10M NaBrO

NaBrO----> Na⁺ + BrO^-

BrO^- + H2O:::::; BrOH + OH^-

Equilibrium concentration

[BrO^-] = 0.10-x

[BrOH] = x

[OH^-]= x

Kb= 4.0x10^-6

Kb= [BrOH][OH-]/[BrO-]

Kb= x²/0.10-x

Since BrO- is weak, 0.10-x is approximately 0.10

x²= 4.0x10^-6x0.10

x= 6.32x10^-4

[OH-]= 6.32 x 10^-4

pOH= -log[OH^-]= -log(6.32x10^-4)

pOH= 3.2

pH= 14-pOH= 14-3.2= 10.8

2. 0.80M NaHS

NaHs----> Na⁺ + HS^-

HS^- + H2O::::::: H2S + OH^-

Equilibrium concentration

[HS^-] = 0.80-x

[H2S] = x

[OH-] = x

Kb= 9.09x10^-8

Kb= [H2S][OH^-]/[HS^-]= x²/0.80-x

Since HS- is weak, 0.80-x is approximately 0.80

x²= kb x 0.80

x= 9.09x10^-8x0.80

x= 2.7 x 10^-4

[OH-]= 2.7x10^-4

pOH= -log(2.7x10^-4)

pOH= 3.57

pH= 14-3.57= 10.43

3. CH3COONa----> CH3COO- + Na+

(CH3COO)2Ba----> 2CH3COO- + Ba2+

Total concentration of CH3COO- = 0.045 + 2(0.055)

= 0.155M

CH3COO- + H2O:::::: CH3COOH + OH-

Equilibrium concentration

[CH3COO-] = 0.155-x

[CH3COOH]= x

[OH-]= x

Kb= [CH3COOH][OH-]/[CH3COO-]

= x²/0.155-x

0.155-x is approximately 0.155

x²= Kb x 0.155

x²= 5.6x10^-10 x 0.155

x= 9.3x10^-6

[OH] =9.3x10^-6

pOH= -log [OH-]

pOH= 5.03

pH= 14-5.03= 8.97