Question

a 20.0 ml sample of 0.10 mol... CAN YOU EXPLAIN How to do B,C,D plz. i need it asap!! thank you so much

A 20.0 mL sample of 0.10 mol/L CH3COOH was titrated with 0.20 mol/L. NaOH. Ka for CH3COOH is 1.8 x 10-6 a. What volume of NaOH is needed to reach the equivalence point? mL b. Calculate the pH of the solution in the flask at the equivalence point. pH = c. Calculate the pH of the solution in the flask at the equivalence point. pH= d. Calculate the pH of the solution in the flask after 15 mL of NaOH have been added pH=

Answer 1

a)

find the volume of NaOH used to reach equivalence point

M(CH3COOH)*V(CH3COOH) =M(NaOH)*V(NaOH)

0.1 M *20.0 mL = 0.2M *V(NaOH)

V(NaOH) = 10 mL

Answer: 10 mL

b)

At half equivalence point, pH = pKa

use:

pKa = -log Ka

= -log (1.8*10^-5)

= 4.7447

So, pH = 4.7447

Answer: 4.74

c)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 20 mL

M(NaOH) = 0.2 M

V(NaOH) = 10 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 10 mL = 2 mmol

We have:

mol(CH3COOH) = 2 mmol

mol(NaOH) = 2 mmol

2 mmol of both will react to form CH3COO- and H2O

CH3COO- here is strong base

CH3COO- formed = 2 mmol

Volume of Solution = 20 + 10 = 30 mL

Kb of CH3COO- = Kw/Ka = 1*10^-14/1.8*10^-5 = 5.556*10^-10

concentration ofCH3COO-,c = 2 mmol/30 mL = 0.0667M

CH3COO- dissociates as

CH3COO- + H2O -----> CH3COOH + OH-

0.0667 0 0

0.0667-x x x

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

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)*6.667*10^-2) = 6.086*10^-6

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

so, x = 6.086*10^-6 M

[OH-] = x = 6.086*10^-6 M

use:

pOH = -log [OH-]

= -log (6.086*10^-6)

= 5.2157

use:

PH = 14 - pOH

= 14 - 5.2157

= 8.7843

Answer: 8.78

d)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 20 mL

M(NaOH) = 0.2 M

V(NaOH) = 15 mL

mol(CH3COOH) = M(CH3COOH) * V(CH3COOH)

mol(CH3COOH) = 0.1 M * 20 mL = 2 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2 M * 15 mL = 3 mmol

We have:

mol(CH3COOH) = 2 mmol

mol(NaOH) = 3 mmol

2 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 20 + 15 = 35 mL

[OH-] = 1 mmol/35 mL = 0.0286 M

use:

pOH = -log [OH-]

= -log (2.857*10^-2)

= 1.5441

use:

PH = 14 - pOH

= 14 - 1.5441

= 12.4559

Answer: 12.46