Chemistry, 30.03.2020 20:20, preciadogabriel40

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

Answers: 3

Show answers

Answers

answered: sierranowe2003

Here is the full question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

a) 10.00 mL

pH =

b) 20.10 mL

pH =

c) 25.00 mL

pH =

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}$

= 0.001000 mol

pKa of butanoic acid = - log Ka

= - log ( 1.54 × 10⁻⁵)

= 4.81

Equation for the reaction is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

The ICE Table is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.001000 0

Change - 0.001000 - 0.001000 + 0.001000

Equilibrium 0.001000 0 0.001000

Total Volume = (20.00 + 10.00 ) mL

= 30.00 mL = 0.03000 L

Concentration of [CH₃CH₂CH₂COOH] = $\frac{0.001000 \ mol}{ 0.03000 \ L }$

= 0.03333 M

Concentration of [CH₃CH₂COO⁻] = $\frac{0.001000 \ mol}{ 0.03000 \ L}$

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log $\frac{conjugate \ base}{acid }$

pH = pKa + log $\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}$

PH = 4.81 + log $\frac{0.03333}{0.03333}$

pH = 4.81

Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81

b )

After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002010 mol

Following the previous equation of reaction , The ICE Table for this process is as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002010 0

Change - 0.002000 -0.002000 + 0.002000

Equilibrium 0 0.000010 0.002000

We can see here that the base is present in excess;

NOW, number of moles of base present in excess

= ( 0.002010 - 0.002000) mol

= 0.000010 mol

Total Volume = (20.00 + 20.10 ) mL

= 40.10 mL × $\frac{1 \ L}{1000 \ mL }$

= 0.04010 L

Concentration of acid [OH⁻] = $\frac{0.000010 \ mol}{0.04010 \ L }$

= $2.494*10^{-4} M$

Using the ionic product of water:

$[H_3O^+] = \frac{K \omega }{[OH^-]}$

where

$K \omega = 10^{-14}$

$[H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}$

= $4.0*10^{-11}M$

pH = - log [H_3O^+}]

pH = - log [ $4.0*10^{-11}M$ ]

pH = 10.40

Thus, the pH of the solution after the equivalence point = 10.40

After the equivalence point, pH of the solution is determined by the excess H⁺.

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002500 mol

From our chemical equation; The ICE Table can be illustrated as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002500 0

Change - 0.002000 -0.002000 +0.002000

Equilibrium 0 0.000500 0.002000

Base is present in excess

Number of moles of base present in excess = [ 0.002500 - 0.002000] mol

= 0.000500 mol

Total Volume = ( 20.00 + 25.00 ) mL

= 45.00 mL

= 45.00 × $\frac{1 \ L}{1000 \ mL }$

= 0.04500 L

Concentration of acid [OH⁻] = $\frac{0.0005000 \ mol}{ 0.04500 \ L }$

= 0.01111 M

Using the ionic product of water $[H_3O^+] = \frac{K \omega }{[OH^+]}$

= $\frac{1.0*10^{-14}}{0.01111}$

= $9.0*10^{-13}$ M

pH = - log [H_3O^+}]

pH = - log [ $9.0*10^{-13}M$ ]

pH = 12.04

Thus, the pH of the solution after the equivalence point = 12.04

answered: Guest

h let me think

answered: Guest

c. is the answer

by love gaming

Other questions on the subject: Chemistry

Chemistry, 21.06.2019 21:00, HelenKellerwasaSlutt

The earth's moon is unusually large. two popular theories of the moon's origin include the "sister world" hypothesis, which states that the moon formed from the same materials as the earth, near enough to the earth that they fell into orbit around each other. a second theory is the "capture" hypothesis, in which the moon formed elsewhere in the solar system, and the earth's gravity pulled it into its orbit. studies of what the moon is made of indicate that some of its materials had to come from the earth or from the same area of the solar system where the earth had formed. at the same time, the moon does not contain much of the material that makes up the earth's core, so the moon could not have formed from the same materials as the earth. how do the two facts above affect the described theories of the moon's origin? a. they show that scientists will never agree on where the moon came from. b. they show that more experiments on moon formation need to be done. c. they show that no theory accounts for the existence of the moon. d. they show that neither theory is complete and entirely correct.

Answers: 2

continue

Chemistry, 22.06.2019 07:30, gwenparks

Calculate the ratio of h+ ions to oh– ions at a ph = 7. find the concentration of h+ ions to oh– ions listed in table b of your student guide. then divide the h+ concentration by the oh– concentration. record this calculated ratio in table a of your student guide. compare your approximated and calculated ratios of h+ ions to oh– ions at a ph = 7. are they the same? why or why not? record your comparison in table a. what is the concentration of h+ ions at a ph = 7? mol/l what is the concentration of oh– ions at a ph = 7? mol/l what is the ratio of h+ ions to oh– ions at a ph = 7? : 1

Answers: 1

continue

Chemistry, 22.06.2019 11:00, jodonw5616

The diagram below shows the different phase transitions that occur in matter. which arrow represents the transition in which dew is formed?

Answers: 1

continue

Chemistry, 22.06.2019 13:00, devontemiles8868

One of the hopes for solving the world's energy problem is to make use of the fusion reaction 21h +31h --> 42he + 10n + energy how much energy is released when 1 mol of deuterium is fused with 1 mol of tritium according to the above reaction? the masses of the atoms and the neutrons are as follows: 21h = 2.0140 amu 31h = 3.01605 amu 42he = 4.002603 amu 10n = 1.008665 amu. the speed of light is 2.9979 x 108 m/s.

Answers: 1

continue

Do you know the correct answer?

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 ×...