1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Masteriza [31]
3 years ago
11

Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop

anoate (Na C3H5O2) (c) 0.1000M HC3H5O2 and 0.1000M (Na C3H5O2
Chemistry
2 answers:
Vesnalui [34]3 years ago
5 0

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:

HC_{3} H_{5}O_{2}⇄H^{+}+C_{3}H_{5} O_{2}^{-}

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:

K_{a}___HC_3H_5O_2=\frac{[H^+][C_3H_5O_2]^-}{[HC_3H_5O_2]} (4)

We should replace the concentrations in the equation (4) with the concentrations in the equilibrium. Eq (1), (2) and (3)

Ka=\frac{x*x}{0.1000-x} (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=\frac{x*x}{0.1}

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}

We found x, so we know [H+]

The equation to find the pH is:

pH= -Log [H^{+} ]\\pH=-Log(1.14x10^{-3} )\\pH=2.94

(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:

C_3H_5O_2^- + H_2O⇄HC_3H_5O_2 + OH^- (6)

In this chemical equation the water donates hydrogen so it behaves like a base . We need basic ionization constant (K_{b_C_3H_5O_2^-})

To find Kb, we have the follow equation:

Kb=\frac{Kw}{Ka} (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):

Kb=\frac{1x10^{-14}}{1.3x10^{-5}}

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=\frac{[OH^-][HC_3H_5O_2]}{[C_3H_5O_2]} (11)

We should replace the concentrations in the equation (11) with the concentrations in the equilibrium

Kb=\frac{x*x}{0.1-x} (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=\frac{x*x}{0.1}

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:

NaC_3H_5O_2→Na^{+} + C_3H_5O_2^{-} (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:

HC_{3} H_{5}O_{2}⇄H^{+}+C_{3}H_{5} O_{2}^{-} (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:

K_{a}___HC_3H_5O_2=\frac{[H^+][C_3H_5O_2]^-}{[HC_3H_5O_2]}

We should replace the concentrations in the last equation with the concentrations in the equilibrium.

Ka=\frac{x*(0.1+x)}{0.1-x} (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=\frac{x*(0.1+x)}{0.1}

Ka*(0.1)=x(0.1+x)\\Ka*0.1=0.1x+x^2\\Ka*0.1-(0.1*x)-x^2=0\\

1.3x10^{-6}-(0.1*x)-x^2=0 (20)

Equation 20 is solved using the quadratic equation:

x=\frac{-b+/-\sqrt{b^2-4ac} }{2a} (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:

pH= -Log [H^{+} ]\\pH=-Log(1.3x10^{-5} )\\pH=4.89

jek_recluse [69]3 years ago
3 0

(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.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is  1.3 x 10^{-5} .

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:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

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.

You might be interested in
A 0.80 L sample of gas has a temperature of 27°C and a pressure of 0.925 atm. How many moles of gas are present?
ololo11 [35]

Answer:

n= 0.03 moles

Explanation:

Using the ideal gas law:

PV=nRT

nRT=PV

n= PV/RT

 n: moles

 P: pressure in atm

 V= volume in L

 R= Avogadro's constant = 0.0821

 T= Temperature in K => ºC+273.15

n= (0.925 atm)(0.80 L) / (0.0821)(300.15 K)

n= 0.03 moles

6 0
3 years ago
5. A sample of gas occupies 1.55 L at 27 degrees Celsius and 1.00 atm pressure. What will the volume be if the pressure is incre
Sergio039 [100]

Answer:

The volume will be 0.031 L

Explanation:

Since temperature is constant, Boyle's law is applied in this case.

Boyle's law States that at constant temperature, the volume of a given mass of a gas is inversely proportional to it's pressure.

From this statement it was deduced that P1V1 = P2V2

From the question, P1 = 1atm, V1 = 1.55L, P2 = 50 atm and V2 is to be calculated.

V2 = P1V1/P2

= 1×1.55÷50

=0.031 L

5 0
3 years ago
2.93 mol of MgF2 to grams
mojhsa [17]

The mass of 1.72 mol of magnesium fluoride is 107 grams.

To determine the mass of 1.72 mol of magnesium fluoride, we first need the chemical formula of magnesium fluoride. Magnesium forms a +2 ion (Mg+2) and fluoride forms a -1 ion (F-1). Since all compounds formed from ions have to be electrically neutral, we need 2 fluoride ions and 1 magnesium ion. Therefore, the formula for magnesium fluoride is MgF2.

Now we need to determine the molar mass of the compound from the molar mass values from the periodic table. Let's use a table to calculate this molar mass.

Molar mass of MgF2

Element Molar Mass (g/mol) Quantity Total (g/mol)

Mg 24.31 1 24.31

F 19.00 2 38.00

Total molar mass of MgF2 = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol

This is the mass of one mole of the substance. If we have 1.72 mols of it, we multiply 1.72 by 62.31.

1.72 mol (62.31 g/mol) = 107 grams

We rounded to 107 to keep the correct number of significant digits in our answer.

8 0
2 years ago
A theater director reinterprets a play by replacing its classical score with
cricket20 [7]

Answer:

The play will be more  appealing to a younger audience.

Explanation:

A younger audience will more likely appreciate current pop hits rather than classical score.

4 0
2 years ago
Read 2 more answers
Which of the following is the most common gas in the troposphere
Effectus [21]
The most common gas in the troposphere is nitrogen. 

I hope this helps!
7 0
3 years ago
Read 2 more answers
Other questions:
  • A plot of velocity versus substrate concentration for a simple enzyme-catalyzed reaction produces a _____. This indicates that a
    11·1 answer
  • Lollll I need help with this question
    9·1 answer
  • Which of these is an example of formation of a substance with a different identity
    5·2 answers
  • Q4 describe the change of atomic radio for the elements in period 2, from lithium to neon
    10·1 answer
  • What is the concentration of Agt in a 1.2 x 10-4 solution of Ag2CO3? (To write your answer using scientific notation use 1.0E-1
    9·1 answer
  • Directions. Carefully review the question. Use the following choices to answer 1 - 10.
    12·1 answer
  • ¿que fuerza de debera aplicar en un embolo pequeño de una prensa hidraulica de 12cm2 de area ,para levantar un cuerpo de 70000N
    13·1 answer
  • To make an object start moving, you must exert a force in
    7·1 answer
  • 6. The central selenium atom in selenium hexafluo-
    15·1 answer
  • The two camps have the same schedule, the same activities, even the same-looking food. What are all these variables called?​
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!