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
blondinia [14]
3 years ago
7

6. A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCI. Calculate the pH

Chemistry
1 answer:
Vadim26 [7]3 years ago
5 0

Answer:

a) pH = 9.14

b) pH = 8.98

c) pH = 8.79

Explanation:

In this case we have an acid base titration. We have a weak base in this case the pyridine (C₅H₅N) and a strong acid which is the HCl.

Now, we want the know the pH of the resulting solution when we add the following volume of acid: 0, 10 and 20.

To know this, we first need to know the equivalence point of this titration. This can be known using the following expression:

M₁V₁ = M₂V₂  (1)

Using this expression, we can calculate the volume of acid required to reach the equivalence point. Doing that we have:

M₁V₁ = M₂V₂

V₁ = M₂V₂ / M₁

V₁ = 0.125 * 25 / 0.1 = 31.25 mL

This means that the acid and base will reach the equivalence point at 31.25 mL of acid added. So, the volume of added acid of before, are all below this mark, so we can expect that the pH of this solution will be higher than 7, in other words, still basic.

To know the value of pH, we need to apply the following expression:

pH = 14 - pOH  (2)

the pOH can be calculated using this expression:

pOH = -log[OH⁻]  (3)

The [OH⁻] is a value that can be calculated when the pyridine is dissociated into it's ion. However, as this is a weak acid, the pyridine will not dissociate completely in solution, instead, only a part of it will be dissociated. Now, to know this, we need the Kb value of the pyridine.

The reported Kb value of the pyridine is 1.5x10⁻⁹ so, with this value we will do an ICE chart for each case, and then, calculate the value of the pH.

<u>a) 0 mL of acid added.</u>

In this case, the titration has not begun, so the concentration of the base will not be altered. Now, with the Kb value, let's write an ICE chart to calculate the [OH⁻], the pOH and then the pH:

       C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.125                                0             0

e)        -x                                   +x           +x

c)      0.125-x                              x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.125-x --> Kb is really small, so we can assume that x will be very small too, and 0.125-x can be neglected to only 0.125, and then:

1.5x10⁻⁹ = x² / 0.125

1.5x10⁻⁹ * 0.125 = x²

x = [OH⁻] = 1.37x10⁻⁵ M

Now, we can calculate the pOH:

pOH = -log(1.37x10⁻⁵) = 4.86

Finally the pH:

pH = 14 - 4.86

<h2>pH = 9.14</h2>

<u>b) 10 mL of acid added</u>

In this case the titration has begun so the acid starts to react with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.010) = 1x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 1x10⁻³ = 2.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 2.125x10⁻³ / (0.025 + 0.010) = 0.0607 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.0607                             0             0

e)        -x                                   +x           +x

c)      0.0607-x                           x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.0607-x --> 0.0607

1.5x10⁻⁹ = x² / 0.0607

1.5x10⁻⁹ * 0.0607 = x²

x = [OH⁻] = 9.54x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(9.54x10⁻⁶) = 5.02

Finally the pH:

pH = 14 - 5.02

<h2>pH = 8.98</h2>

<u>c) 20 mL of acid added:</u>

In this case the titration it's almost reaching the equivalence point and the acid is still reacting with the base, so we need to know how many moles of the base remains after the volume of added acid:

moles acid = 0.1 * (0.020) = 2x10⁻³ moles

moles base = 0.125 * 0.025 = 3.125x10⁻³

This means that the base is still in higher quantities, and the acid is the limiting reactant here, so the remaining moles will be:

remaining moles of pyridine = 3.125x10⁻³ - 2x10⁻³ = 1.125x10⁻³ moles

The concentration of pyridine in solution:

[C₅H₅N] = 1.125x10⁻³ / (0.025 + 0.020) = 0.025 M

Now with this concentration, we will do the same procedure of before, with the ICE chart, but replacing this new value of the base, to get the [OH⁻] and then the pH:

        C₅H₅N + H₂O <-------> C₅H₅NH⁺ + OH⁻     Kb = 1.5x10⁻⁹

i)       0.025                                0             0

e)        -x                                   +x           +x

c)      0.025-x                             x             x

Writting the Kb expression:

Kb = [C₅H₅NH⁺] [OH⁻] / [C₅H₅N]    replacing the values of the chart:

1.5x10⁻⁹ = x² / 0.025-x --> 0.025

1.5x10⁻⁹ = x² / 0.025

1.5x10⁻⁹ * 0.025 = x²

x = [OH⁻] = 6.12x10⁻⁶ M

Now, we can calculate the pOH:

pOH = -log(6.12x10⁻⁶) = 5.21

Finally the pH:

pH = 14 - 5.21

<h2>pH = 8.79</h2>
You might be interested in
Density is considered what type of property
Taya2010 [7]
Density is a physical property. It's measured and doesn't change the object chemically.
4 0
3 years ago
What is △n for the following equation in relating Kc to Kp?
Nimfa-mama [501]

Answer:

-1  

Explanation:

The relation between Kp and Kc is given below:

K_p= K_c\times (RT)^{\Delta n}

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

2Na_{(s)}+2H_2O_{(l)}\rightleftharpoons 2NaOH_{(aq)}+2H_2_{(g)}

<u>Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)  = (2+1)-(2+2) = -1  </u>

<u></u>

7 0
3 years ago
Name the element which lies in group IA and 3rd period. What is the valency of it and why?​
Jlenok [28]

Answer:

Bro I got notification of this question but soory I don't know the answer Hope you understand me

4 0
2 years ago
Read 2 more answers
In this experiment, 0.070 g of caffeine is dissolved in 4.0 ml of water. The caffeine is then extracted from the aqueous solutio
Damm [24]

2.0ml of methylene chloride solution is used each time to extract caffeine from the aqueous solution.  

Consider the concentration of caffeine obtained during each individual extraction from the aqueous solution to be C.  

The total amount of caffeine obtained during each extraction is calculated as

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution) + (Volume of methylene chloride added during each extraction x distribution coefficient of caffeine x concentration of caffeine obtained during each individual extraction from the aqueous solution)  


Substituting these values we get                                                            

The total amount of caffeine obtained during each extraction                

 = (4.0×C )+ (2.0×4.6 × C)                                                                              

= 13.2 C


The amount of caffeine remaining in the aqueous solution is calculated as  

(Total volume of water used to make up the caffeine aqueous solution) x (concentration of caffeine obtained during each individual extraction from the aqueous solution)


Substituting these values we get                                                            

The amount of caffeine remaining in the aqueous solution = 4 × C                                                                                            

The fraction of caffeine remaining in aqueous solution is calculated as  

= (The total amount of caffeine obtained during each extraction)/ (The amount of caffeine remaining in the aqueous solution)                    

=4.0 C/13.2 C                                                                                                

= 1/3.3.  

Therefore the fraction of caffeine left in aqueous solution after 3 extractions is =(1/3.3)^3  =0.028

Therefore, the total amount of caffeine extracted                            

=0.070 × (1-(1/3.3)^3)                                                                                      

= 0.068 g


5 0
3 years ago
Read 2 more answers
For the reaction 2 nh3 + ch3oh → products, how much ch3oh is needed to react with 93.5 g of nh3? 1. 1.31 mol 2. 46.8 mol 3. 2.75
DENIUS [597]
3.2.75 mol is the answer.
6 0
3 years ago
Read 2 more answers
Other questions:
  • What is pollution ? briefly<br><br>​
    11·1 answer
  • nickel(ii) carbonate + hydrobromic acid = nickel(ii) bromide + carbonic acid + bromine gas how do you balance this equation
    7·1 answer
  • What is the total anion concentration (in mEq/L ) of a solution that contains 6.0 mEq/L Na + , 11.0 mEq/L Ca 2+ , and 1.0 mEq/L
    10·1 answer
  • Where else do you see Force &amp; Motion in real-world experiences?
    6·2 answers
  • You are driving a Winnebago at a casual speed of 54 miles per hour (mph) on a journey to visit your grandmother in Alabama.
    6·1 answer
  • An atom bonds with another atom. What is the best classification for this reaction?
    5·2 answers
  • Please help I will mark brainliest
    8·1 answer
  • Explain how deep ocean currents form.
    9·1 answer
  • What is the basic unit of structure and function in all living things? a cell b organism c atom d plasma
    11·1 answer
  • The atomic number of an element:
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!