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
blagie [28]
2 years ago
14

A 20.0 mL 0.100 M solution of lactic acid is titrated with 0.100 M NaOH.

Chemistry
1 answer:
yan [13]2 years ago
7 0

Answer:

(a) See explanation below

(b) 0.002 mol

(c) (i) pH = 2.4

(ii) pH = 3.4

(iii) pH = 3.9

(iv) pH = 8.3

(v) pH = 12.0

Explanation:

(a) A buffer solution exits after addition of 5 mL of NaOH  since after reaction we will have  both the conjugate base lactate anion and unreacted weak  lactic acid present in solution.

Lets call lactic acid HA, and A⁻ the lactate conjugate base. The reaction is:

HA + NaOH ⇒ A⁻ + H₂O

Some unreacted HA will remain in solution, and since HA is a weak acid , we will have the followin equilibrium:

HA  + H₂O ⇆ H₃O⁺ + A⁻

Since we are going to have unreacted acid, and some conjugate base, the buffer has the capacity of maintaining the pH in a narrow range if we add acid or base within certain limits.

An added acid will be consumed by the conjugate base A⁻ , thus keeping the pH more or less equal:

A⁻ + H⁺ ⇄ HA

On the contrary, if we add extra base it will be consumed by the unreacted lactic acid, again maintaining the pH more or less constant.

H₃O⁺ + B ⇆ BH⁺

b) Again letting HA stand for lactic acid:

mol HA =  (20.0 mL x  1 L/1000 mL) x 0.100 mol/L = 0.002 mol

c)

i) After 0.00 mL of NaOH have been added

In this case we just have to determine the pH of a weak acid, and we know for a monopric acid:

pH = - log [H₃O⁺] where  [H₃O⁺] = √( Ka [HA])

Ka for lactic acid = 1.4 x 10⁻⁴  ( from reference tables)

[H₃O⁺] = √( Ka [HA]) = √(1.4 x 10⁻⁴ x 0.100) = 3.7 x 10⁻³

pH = - log(3.7 x 10⁻³) = 2.4

ii) After 5.00 mL of NaOH have been added ( 5x 10⁻³ L x 0.1 = 0.005 mol NaOH)

Now we have a buffer solution and must use the Henderson-Hasselbach equation.

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.0005                0

after rxn    0.002-0.0005                  0                  0.0005

                        0.0015

Using Henderson-Hasselbach equation :

pH = pKa + log [A⁻]/[HA]

pKa HA = -log (1.4 x 10⁻⁴) = 3.85

pH = 3.85 + log(0.0005/0.0015)

pH = 3.4

iii) After 10.0 mL of NaOH have been ( 0.010 L x 0.1 mol/L = 0.001 mol)

                             HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.001               0

after rxn        0.002-0.001                  0                  0.001

                        0.001

pH = 3.85 + log(0.001/0.001)  = 3.85

iv) After 20.0 mL of NaOH have been added ( 0.002 mol )

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn         0.002                  0.002                 0

after rxn                 0                         0                   0.002

We are at the neutralization point and  we do not have a buffer anymore, instead we just have  a weak base A⁻ to which we can determine its pOH as follows:

pOH = √Kb x [A⁻]

We need to determine the concentration of the weak base which is the mol per volume in liters.

At this stage of the titration we added 20 mL of lactic acid and 20 mL of NaOH, hence the volume of solution is 40 mL (0.04 L).

The molarity of A⁻ is then

[A⁻] = 0.002 mol / 0.04 L = 0.05 M

Kb is equal to

Ka x Kb = Kw ⇒ Kb = 10⁻¹⁴/ 1.4 x 10⁻⁴ = 7.1 x 10⁻¹¹

pOH is then:

[OH⁻] = √Kb x [A⁻]  = √( 7.1 x 10⁻¹¹ x 0.05) = 1.88 x 10⁻⁶

pOH = - log (  1.88 x 10⁻⁶ ) = 5.7

pH = 14 - pOH = 14 - 5.7 = 8.3

v) After 25.0 mL of NaOH have been added (

                            HA          +         NaOH          ⇒   A⁻ + H₂O

before rxn           0.002                  0.0025              0

after rxn                0                         0.0005              0.0005

Now here what we have is  the strong base sodium hydroxide and A⁻ but the strong base NaOH will predominate and drive the pH over the weak base A⁻.

So we treat this part as the determination of the pH of a strong base.

V= (20 mL + 25 mL) x 1 L /1000 mL = 0.045 L

[OH⁻] = 0.0005 mol / 0.045 L = 0.011 M

pOH = - log (0.011) = 2

pH = 14 - 1.95 = 12

You might be interested in
Write the balanced COMPLETE ionic equation for the reaction when
Vera_Pavlovna [14]

Answer:

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

Explanation:

Let's consider the molecular equation between barium nitrate and rubidium hydroxide to produce barium hydroxide and rubidium nitrate.

Ba(NO₃)₂(aq) + 2 RbOH(aq) = Ba(OH)₂(s) + 2 RbNO₃(aq)

The complete ionic equation includes all the ions and the molecular species.

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

3 0
3 years ago
Which option describes energy being released as heat?
Mnenie [13.5K]

Answer:

I’m pretty sure it’s Lions sleeping after a big meal

Explanation:

8 0
3 years ago
Read 2 more answers
If during an experiment zinc was found to be more reactive than lead or copper, zinc would be considered the strongest _____.
densk [106]

Zinc would be considered the strongest reducing agent.

<h3>Reducing agent</h3>

A reducing agent is a chemical species that "donates" one electron to another chemical species in chemistry (called the oxidizing agent, oxidant, oxidizer, or electron acceptor). Earth metals, formic acid, oxalic acid, and sulfite compounds are a few examples of common reducing agents.

Reducers have excess electrons (i.e., they are already reduced) in their pre-reaction states, whereas oxidizers do not. Usually, a reducing agent is in one of the lowest oxidation states it can be in. The oxidation state of the oxidizer drops while the oxidizer's oxidation state, which measures the amount of electron loss, increases. The agent in a redox process whose oxidation state rises, which "loses/donates electrons," which "oxidizes," and which "reduces" is known as the reducer or reducing agent.

Learn more about reducing agent here:

brainly.com/question/2890416

#SPJ4

<h3 />
5 0
1 year ago
The following balanced equation shows the decomposition of ammonia (NH3) into nitrogen (N2) and hydrogen (H2). 2NH3 → N2 + 3H2 A
Triss [41]
The decomposition of ammonia is characterized by the following decomposition equation:
                                  2NH₃<span>   →   N</span>₂  <span> +   3H</span>₂   

The mole ratio of N₂  :  H₂  is  1  :  3

    If the number of moles of N₂  =  0.0351 mol
    Then the number of moles of H₂  =  0.0351 mol  × 3
                                                         = 0.1053 mol

The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is  0.105 mol (OPTION 3).

6 0
3 years ago
Read 2 more answers
Onsider the following reaction at equilibrium:
11111nata11111 [884]

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

C(s)+H_2O(l)\leftrightharpoons CO(g)+H_2(g)

A. C is added to the reaction mixture.

If the concentration of reactant specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of reactant specie occurs. So, on increasing C ,equilibrium will shift in right direction.

B. H_2O is condensed and removed from the reaction mixture.

If the concentration of reactant specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of reactant specie occurs. So, on removing water vapor ,the equilibrium will shift in left direction.

C. CO is added to the reaction mixture.

If the concentration of product specie is increased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where decrease in concentration of product specie occurs. So, on increasing CO, the equilibrium will shift in left direction.

D. H_2 is removed from the reaction mixture.

If the concentration of product specie is decreased , according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in concentration of product specie occurs. So, on removing hydrogen , the equilibrium will shift in right direction.

5 0
3 years ago
Other questions:
  • Which of the following is not a part of Darwin’s theory of evolution?
    9·2 answers
  • Mac mixed 24 g of Substance A and 12 g of Substance B to form 10 g of Substance C and a certain amount of Substance D. Which sta
    5·2 answers
  • What does the R stand for in the ideal gas law (PV=nRT)?
    8·1 answer
  • What is the Molecule Geometry of a ABE3 molecule?
    14·1 answer
  • What is the most important clue that a chemical reaction is taking place?
    15·1 answer
  • Magnesium oxidizes via the reaction: 2 Mg + O2 → 2 MgO The reaction has a △Hrxn = -1203 kJ. How much heat (in kJ) is released wh
    10·1 answer
  • 1) What is bioluminescence? ​
    5·1 answer
  • A balloon is filled with 3.50 L of water at 24.0 C and 2.27 atm. If the balloon is placed outdoors on a hot day at a temperature
    14·1 answer
  • What is the oxidation state of C in NaHCO3?<br> O A. +4<br> O B. -6<br> O C. +1<br> OD. -2
    14·1 answer
  • Sample Data - Grignard Reagents Reaction Characterization Type of reaction: choose the general type of reaction from the options
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!