Question

can you please answer 1,2, and 3 and show the steps

1. Calculate the pH of a buffer solution made by adding 20.0 mL of 0.200 M acetic acid solution with 10.0 mL of a 0.200 solution of sodium acetate. K, for acetic acid is 1.8 x 109. Assume the total volume is 30.0 mL. 2. Calculate the pH of a buffer solution made by mixing 10.0 mL of 0.20 M ammonia with 15.0 mL of 0,15 M ammonium chloride solution. K, for ammonia is 1.8 x 10-5. Assume the total volume is 25.0 mL. 3. Calculate the pH of a 0.15 M sodium formate solution. K, for formic acid is acid is 1.8 x 10 4. Calculate the molar solubility of lead chloride in 0.10 M sodium chloride solution. K for lead chloride is acid is 1.2 x 105. 5. Calculate the concentration of the FeNCS2+ complex ion in a solution obtained by adding 10.0 mL of a 0.080 M Fe3+ ion solution to 10.0 mL of a 0.00200 M NCS ion solution. K. = 450. 6. The Fe3+ ion solution in Q. 5 above is usually prepared in a sulfuric acid solution. Why?

Answer 1

1)

Concentration after mixing = mol of component / (total volume)

M(CH3COO-) after mixing = M(CH3COO-)*V(CH3COO-)/(total volume)

M(CH3COO-) after mixing = 0.2 M*10.0 mL/(10.0+20.0)mL

M(CH3COO-) after mixing = 6.667*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(CH3COOH) after mixing = M(CH3COOH)*V(CH3COOH)/(total volume)

M(CH3COOH) after mixing = 0.2 M*20.0 mL/(20.0+10.0)mL

M(CH3COOH) after mixing = 0.1333 M

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.745+ log {6.667*10^-2/0.1333}

= 4.444

Answer: 4.44

2)

Concentration after mixing = mol of component / (total volume)

M(NH3) after mixing = M(NH3)*V(NH3)/(total volume)

M(NH3) after mixing = 0.2 M*10.0 mL/(10.0+15.0)mL

M(NH3) after mixing = 8*10^-2 M

Concentration after mixing = mol of component / (total volume)

M(NH4+) after mixing = M(NH4+)*V(NH4+)/(total volume)

M(NH4+) after mixing = 0.15 M*15.0 mL/(15.0+10.0)mL

M(NH4+) after mixing = 9*10^-2 M

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 4.745+ log {9*10^-2/8*10^-2}

= 4.796

use:

PH = 14 - pOH

= 14 - 4.7959

= 9.2041

Answer: 9.20

3)

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.8*10^-4

Kb = 5.556*10^-11

HCOO- dissociates as

HCOO- + H2O -----> HCOOH + OH-

0.15 0 0

0.15-x x x

Kb = [HCOOH][OH-]/[HCOO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-11)*0.15) = 2.887*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.887*10^-6 M

use:

pOH = -log [OH-]

= -log (2.887*10^-6)

= 5.5396

use:

PH = 14 - pOH

= 14 - 5.5396

= 8.4604

Answer: 8.46