Question

Can I get help with 2,3, and 4.

(pH = 2.72 2. Calculate the pH of a 0.10 M aqueous solution of sodium acetate, NaC2H302. (2 pts) 3. A buffer solution contains 0.50 M acetic acid and 0.50 sodium acetate. Calculate the pH of this solution. (2 pts) 4. What is the pH of 2.00mL of 1.0 M HCl solution diluted to 50.0 mL? ( 2pts)

Answer 1

2)

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^-5

Kb = 5.556*10^-10

C2H3O2- dissociates as

C2H3O2- + H2O -----> HC2H3O2 + OH-

0.1 0 0

0.1-x x x

Kb = [HC2H3O2][OH-]/[C2H3O2-]

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^-10)*0.1) = 7.454*10^-6

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

so, x = 7.454*10^-6 M

use:

pOH = -log [OH-]

= -log (7.454*10^-6)

= 5.1276

use:

PH = 14 - pOH

= 14 - 5.1276

= 8.8724

Answer: 8.87

3)

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 {0.5/0.5}

= 4.745

Answer: 4.75

4)

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 1 M

V1 = 2 mL

V2 = 50 mL

use:

M1*V1 = M2*V2

M2 = (M1*V1)/V2

M2 = (1*2)/50

M2 = 0.040 M

Answer: 0.040 M