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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 = <u> </u>
b) 20.10 mL
pH = <u> </u>
c) 25.00 mL
pH = <u> </u>
<u />
Answer:
pH = 4.81
pH = 10.40
pH = 12.04
Explanation:
a)
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 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] =
= 0.03333 M
Concentration of [CH₃CH₂COO⁻] =
= 0.03333 M
By Henderson- Hasselbalch equation
pH = pKa + log
pH = pKa + log
PH = 4.81 + log
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
=
= 0.002000 mol
Number of moles of NaOH added
=
= 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 ×
= 0.04010 L
Concentration of acid [OH⁻] =
=
Using the ionic product of water:
where
=
pH = - log
pH = - log []
pH = 10.40
Thus, the pH of the solution after the equivalence point = 10.40
c)
After the equivalence point, pH of the solution is determined by the excess H⁺.
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 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 ×
= 0.04500 L
Concentration of acid [OH⁻] =
= 0.01111 M
Using the ionic product of water
=
= M
pH = - log
pH = - log []
pH = 12.04
Thus, the pH of the solution after the equivalence point = 12.04
The answer is (A.) 135.6 N
————
Here is the formula you will need to answer the question.
Formula:
Force(N) = Mass(kg) x Acceleration(m/s²)
or... F(N) = m(kg) x a(m/s²)
————
We need to be certain of the information presented. The problem says the man has a "Mass" of 83 kilograms on Earth. If a man has a Mass of 83 kilograms on Earth, he has a mass of 83 kilograms everywhere. The mass for this problem (even on the moon) is 83.
Mass(kg): 83
————
To find the acceleration of object on other planets, moons, ect, you'll need to search the internet or ask a teacher. There are ways to calculate this, but the problem didn't supply the necessary information to do so.
Acceleration: 1.625 m/s²
————
Now just plug the information into the formula.
F(N) = m(kg) x a(m/s²)
N = 83 x 1.625
N= 135.875
When considering that there could be a slighty more accurate number for acceleration, it can be concluded that the best answer is (A.)
————
Here is the formula you will need to answer the question.
Formula:
Force(N) = Mass(kg) x Acceleration(m/s²)
or... F(N) = m(kg) x a(m/s²)
————
We need to be certain of the information presented. The problem says the man has a "Mass" of 83 kilograms on Earth. If a man has a Mass of 83 kilograms on Earth, he has a mass of 83 kilograms everywhere. The mass for this problem (even on the moon) is 83.
Mass(kg): 83
————
To find the acceleration of object on other planets, moons, ect, you'll need to search the internet or ask a teacher. There are ways to calculate this, but the problem didn't supply the necessary information to do so.
Acceleration: 1.625 m/s²
————
Now just plug the information into the formula.
F(N) = m(kg) x a(m/s²)
N = 83 x 1.625
N= 135.875
When considering that there could be a slighty more accurate number for acceleration, it can be concluded that the best answer is (A.)