Answer :
(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.
(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.
(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.
Further explanation:
(a)
Given information:
The value of acid ionization constant for propanoic acid is 1.3 x [tex]10^{-5}[/tex] .
The initial concentration of propanoic acid is .
To calculate:
The pH of 0.1000 M propanoic acid solution.
Solution:
Propanoic acid is a weak acid. It ionizes partially in water as follows:
The expression for acid dissociation constant is,
…… (1)
Here,
is ionization constant of propanoic acid.
is the equilibrium concentration of propanoate ion.
is the equilibrium concentration of hydronium ion.
is the equilibrium concentration of propanoic acid.
ICE table (1):
Refer ICE table (1),
Substitute the values form the ICE table (1) in equation (1).
The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,
Rearrange above equation for x.
…… (2)
Substitute for in equation (2) to calculate the value of x.
Therefore, from the ICE table (1) the concentration of hydronium ion is,
.
The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,
…… (3)
Substitute for in equation (3) to calculate the pH of the solution.
(b)
Given information:
The value of acid ionization constant for propanoic acid is .
The initial concentration of sodium propanoate is .
To calculate:
The pH of 0.1000 M sodium propanoate solution.
Solution:
Sodium propanoate is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:
…… (4)
Propanoic acid is a weak acid. It ionizes partially in water as follows:
…… (5)
Dissociation reaction for water is written as follows:
…… (6)
From equation (4), (5), and (6) the relationship between and is,
…… (7)
Substitute for and for in equation (7).
ICE table (2):
The expression for base dissociation constant is,
…… (8)
Here,
is base ionization constant.
is the equilibrium concentration of propanoate ion.
is the equilibrium concentration of hydroxide ion.
is the equilibrium concentration of propanoic acid.
From the ICE table (2),
Substitute the values form the ICE table (2) in equation (8).
The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,
Rearrange above equation for y.
…… (9)
Substitute for in equation (9) to calculate the value of y.
Therefore, from the ICE table (2) the concentration of hydroxide ion is,
The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,
…… (10)
Substitute for in equation (10) to calculate pOH of the solution.
The relation between pH and pOH is as follows:
…… (11)
Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.
(c)
Given information:
The value of acid ionization constant for propanoic acid is .
The initial concentration of sodium propanoate is .
The initial concentration of sodium propanoate is .
To calculate:
The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.
Solution:
Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.
The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,
For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,
…… (12)
The negative logarithm of acid ionization constant is equal to .
…… (13)
Substitute for in equation (13).
Substitute for , for and 4.9 for in equation (12).
Learn more:
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Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Ionic equilibria
Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.
Answer:
(a) 0.1000 M Propanoic acid, Ka=1.3x10-5 pH=2.94
(b) 0.1000M sodium propanoate, pH=8.94
(c) 0.1000M HC3H5O2 and 0.100M NaC3H5O2 pH=4.89
Explanation:
(a) Propanoic acid is a weak acid, the general equation is:
[tex]HC_{3} H_{5}O_{2}[/tex]⇄[tex]H^{+}+C_{3}H_{5} O_{2}^{-}[/tex]
At the beginning of the reaction, the initial concentration are:
HC3H5O2=0.1
H+=0
C3H5O2=0
After the beginning of the reaction, the concentrations change:
HC3H5O2= -x
H+= x
C3H5O2= x
When the reaction reaches the equilibrium, the final concentrations will be the initial concentrations minus the change concentrations:
HC3H5O2= 0.1 -x (1)
H+= x (2)
C3H5O2= x (3)
We need the equation of acid ionization constant:
[tex]K_{a}___HC_3H_5O_2[/tex]=[tex]\frac{[H^+][C_3H_5O_2]^-}{[HC_3H_5O_2]}[/tex] (4)
We should replace the concentrations in the equation (4) with the concentrations in the equilibrium. Eq (1), (2) and (3)
Ka=[tex]\frac{x*x}{0.1000-x}[/tex] (5)
As this acid dissociates in very low concentrations, we use the approximation x = 0
Therefore we assume that 0.1 - x will be approximately equal to 0.1
Ka=[tex]\frac{x*x}{0.1}[/tex]
[tex]Ka= \frac{x^2}{0.1}\\Ka*0.1000= {x^2}\\x= \sqrt{Ka*0.1} \\x= \sqrt{1.3x10x^{-5} *0.1} \\x=\sqrt{1,3x10x^{-6} } \\x= 1.14x10x^{-3}[/tex]
We found x, so we know [H+]
The equation to find the pH is:
[tex]pH= -Log [H^{+} ]\\pH=-Log(1.14x10^{-3} )\\pH=2.94[/tex]
(b) Sodium propanoate is a salt. The sodium propanoate comes from a strong base (NaOH) and a weak acid HC3H5O2
The hydrolysis reaction for the anion is:
[tex]C_3H_5O_2^- + H_2O[/tex]⇄[tex]HC_3H_5O_2 + OH^-[/tex] (6)
In this chemical equation the water donates hydrogen so it behaves like a base . We need basic ionization constant ([tex]K_{b_C_3H_5O_2^-}[/tex])
To find Kb, we have the follow equation:
[tex]Kb=\frac{Kw}{Ka}[/tex] (7)
We know the water constant (Kw=1x10-14) and we have the acid ionization constant if Propanoic acid (Ka= 1.3x10-5)
Using the equation (7):
[tex]Kb=\frac{1x10^{-14}}{1.3x10^{-5}}[/tex]
Kb=7.69x10-10
In the hydrolysis reaction (6) , the initial concentrations are:
C3H5O2= 0.1
OH- = 0
HC3H5O2 = 0
After the beginning of the reaction, the concentrations change:
C3H5O2= -x
OH- = x
HC3H5O2 = x
When the reaction reaches the equilibrium, the final concentrations are:
C3H5O2= 0.1 - x (8)
OH- = x (9)
HC3H5O2 = x (10)
We need the equation of basic ionization constant, this equation is:
Kb=[tex]\frac{[OH^-][HC_3H_5O_2]}{[C_3H_5O_2]}[/tex] (11)
We should replace the concentrations in the equation (11) with the concentrations in the equilibrium
Kb=[tex]\frac{x*x}{0.1-x}[/tex] (12)
As this salt dissociates in very low concentrations, we use the approximation x = 0
Therefore we assume that 0.1 - x will be approximately equal to 0.1
so the solution of the equation (12) is:
Kb=[tex]\frac{x*x}{0.1}[/tex]
Kb= \frac{x^2}{0.1}\\Kb*0.1= {x^2}\\x= \sqrt{Kb*0.1} \\x= \sqrt{7.69x10^{-10} *0.1} \\x= 8.77x10x^{-6}[/tex]
We found x, so we know OH-
The equation to find the pOH is:
pOH= -Log [OH-]
pOH= -Log(8.77x10-6)
pOH=5.06
To find the pH we know:
pH + pOH = 14 (13)
pH = 8.94
(c) We have Sodium propanoate and propanoic acid to 0.1M
We need 2 chemical reactions, when the salt dissociates:
[tex]NaC_3H_5O_2[/tex]→[tex]Na^{+} + C_3H_5O_2^{-}[/tex] (14)
The initial concentrations are:
NaC3H5O2=0.1
Na+=0
C3H5O2=0
At the end of the reaction (14), the concentrations are:
NaC3H5O2= 0 (15)
Na+= 0.1 (16)
C3H5O2= 0.1 (17)
The second chemical reaction is:
[tex]HC_{3} H_{5}O_{2}[/tex]⇄[tex]H^{+}+C_{3}H_{5} O_{2}^{-}[/tex] (18)
At the beginning of the reaction, the initial concentration are:
HC3H5O2=0.1
H+=0
C3H5O2=0.1 (This is the final concentration when the salt was dissociated eq. (17))
After the beginning of the reaction, the concentrations change:
HC3H5O2= -x
H+= x
C3H5O2= x
When the reaction reaches the equilibrium, the final concentrations are:
HC3H5O2=0.1 - x
H+= x
C3H5O2=0.1 + x
We need the equation of acid ionization constant, this equation is:
[tex]K_{a}___HC_3H_5O_2[/tex]=[tex]\frac{[H^+][C_3H_5O_2]^-}{[HC_3H_5O_2]}[/tex]
We should replace the concentrations in the last equation with the concentrations in the equilibrium.
Ka=[tex]\frac{x*(0.1+x)}{0.1-x}[/tex] (19)
As this acid dissociates in very low concentrations, we use the approximation x = 0
Therefore we assume that 0.1 - x will be approximately equal to 0.1
Ka=[tex]\frac{x*(0.1+x)}{0.1}[/tex]
[tex]Ka*(0.1)=x(0.1+x)\\Ka*0.1=0.1x+x^2\\Ka*0.1-(0.1*x)-x^2=0\\[/tex]
[tex]1.3x10^{-6}-(0.1*x)-x^2=0[/tex] (20)
Equation 20 is solved using the quadratic equation:
[tex]x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex] (21)
Using the equation (21) we know the coefficients are:
a= -1
b= -0.1000
c= 1.3x10-6
The solution to the equation (21):
x1= - 0.1
x2= 1.3x10-5
We know the concentrations should be positives so the correct answer is 1.3x10-5
Now we know [H+]
The equation to find the ph is:
[tex]pH= -Log [H^{+} ]\\pH=-Log(1.3x10^{-5} )\\pH=4.89[/tex]
