Question

Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3COOH(aq) with 0.2000 M NaOH(aq) after 2.5 mL of the base have been added. Ka of acetic acid = 1.8 x 10-5.

QUESTION 9

Calculate the pH during the titration of 30.00 mL of 0.1000 M ethylamine, C2H5NH2(aq), with 0.1000 M HCl(aq) after 26 mL of the acid have been added. Kb of ethylamine = 6.5 x 10-4.

QUESTION 10

Calculate the pH during the titration of 20.00 mL of 0.1000 M HCOOH(aq) with 0.1000 M NaOH(aq) after 19 mL of the base have been added. Ka of formic acid = 1.8 x 10-4.

Answer 1

8)

Given:

M(CH3COOH) = 0.1 M

V(CH3COOH) = 20 mL

M(NaOH) = 0.2 M

V(NaOH) = 2.5 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 * 2.5 mL = 0.5 mmol

We have:

mol(CH3COOH) = 2 mmol

mol(NaOH) = 0.5 mmol

0.5 mmol of both will react

excess CH3COOH remaining = 1.5 mmol

Volume of Solution = 20 + 2.5 = 22.5 mL

[CH3COOH] = 1.5 mmol/22.5 mL = 0.0667M

[CH3COO-] = 0.5/22.5 = 0.0222M

They form acidic buffer

acid is CH3COOH

conjugate base is CH3COO-

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 {2.222*10^-2/6.667*10^-2}

= 4.268

Answer: 4.27

9)

Given:

M(HCl) = 0.1 M

V(HCl) = 26 mL

M(C2H5NH2) = 0.1 M

V(C2H5NH2) = 30 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 26 mL = 2.6 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HCl) = 2.6 mmol

mol(C2H5NH2) = 3 mmol

2.6 mmol of both will react

excess C2H5NH2 remaining = 0.4 mmol

Volume of Solution = 26 + 30 = 56 mL

[C2H5NH2] = 0.4 mmol/56 mL = 0.0071 M

[C2H5NH3+] = 2.6 mmol/56 mL = 0.0464 M

They form basic buffer

base is C2H5NH2

conjugate acid is C2H5NH3+

Kb = 6.5*10^-4

pKb = - log (Kb)

= - log(6.5*10^-4)

= 3.187

use:

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

= 3.187+ log {4.643*10^-2/7.143*10^-3}

= 4

use:

PH = 14 - pOH

= 14 - 4

= 10

Answer: 10.0

10)

Given:

M(HCOOH) = 0.1 M

V(HCOOH) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 19 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

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

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

mol(NaOH) = 0.1 M * 19 mL = 1.9 mmol

We have:

mol(HCOOH) = 2 mmol

mol(NaOH) = 1.9 mmol

1.9 mmol of both will react

excess HCOOH remaining = 0.1 mmol

Volume of Solution = 20 + 19 = 39 mL

[HCOOH] = 0.1 mmol/39 mL = 0.0026M

[HCOO-] = 1.9/39 = 0.0487M

They form acidic buffer

acid is HCOOH

conjugate base is HCOO-

Ka = 1.8*10^-4

pKa = - log (Ka)

= - log(1.8*10^-4)

= 3.745

use:

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

= 3.745+ log {4.872*10^-2/2.564*10^-3}

= 5.023

Answer: 5.02