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
alexandr1967 [171]
3 years ago
10

A 35.0-ml sample of 0.150 m acetic acid (ch3cooh) is titrated with 0.150 m naoh solution. calculate the ph after 17.5 ml of base

have been added. the ka for the acetic acid equilibrium is 1.8 x 10-5. chegg
Chemistry
2 answers:
Gekata [30.6K]3 years ago
6 0
When the moles of CH3COOH = volume of CH3COOH * no.of moles of CH3COOH
moles of CH3COOH = 35ml * 0.15 m/1000 =0.00525 mol
moles of NaOH = volume of NaOH*no.of moles of NaOH
                           = 17.5 ml * 0.15/1000 = 0.002625
SO the reaction after add the NaOH:
                           CH3COOH(aq) +OH- (aq) ↔ CH3COO-(aq) +H2O(l)
initial                  0.00525                0                         0
change             - 0.002625         +0.002625     +0.002625
equilibrium      0.002625             0.002625        0.002625
When the total volume = 35ml _ 17.5ml = 52.5ml = 0.0525L
∴[CH3COOH] = 0.002625/0.0525 = 0.05m
and [CH3COO-]= 0.002625/0.0525= 0.05 m
when PKa = -㏒Ka
                 = -㏒1.8x10^-5 = 4.74
by substitution in the following formula:
PH = Pka + ㏒[CH3COO-]/[CH3COOH]
      = 4.74 + ㏒(0.05/0.05) = 4.74
∴PH = 4.74
ycow [4]3 years ago
6 0

5 - log 1.8 or 4.74.

<h3>Further explanation</h3>

Given:

  • A 35.0-ml sample of 0.150 M acetic acid (CH₃COOH) is titrated with 0.150 M NaOH solution.  
  • After 17.5 ml of base have been added.
  • The Ka for the acetic acid equilibrium is 1.8 x 10⁻⁵.

Question:

Calculate the pH of this buffer.

The Process:

Step-1

Let us prepare the mole number of the two reagents.

\boxed{ \ n = MV \ }

Moles of CH₃COOH = \boxed{ 0.150 \ \frac{mol}{L} \times 35.0 \ ml = 5.25 \ mmol \ }

Moles of NaOH = \boxed{ 0.150 \ \frac{mol}{L} \times 17.5 \ ml = 2.625 \ mmol \ }

Step-2

Let use the ICE table.

Balanced reaction:

                   \boxed{ \ CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \ }

Initial:                5.25           2.625                -                  -

Change:           -2.625        -2.625          +2.625        +2.625

Equlibrium:      2.625             -                 2.625          2.625

  • NaOH as a strong base acts as a limiting reagent.
  • The remaining CH₃COOH as a weak acid and CH₃COONa salt form an acidic buffer system.
  • The CH₃COONa salt has valence = 1 according to the number of CH₃⁻ ions as a weak part, i.e., \boxed{ \ CH_3COONa \rightleftharpoons CH_3COO^- + Na^+ \ }
  • CH₃COOH and CH₃COO⁻ are conjugate acid-base pairs.

Step-3

Let us prepare pKa and concentrations of CH₃COOH and CH₃COO⁻.

pK_a = -log(Ka)

\boxed{ \ pK_a = -log[1.8 \cdot 10^{-5}] = 5 - log \ 1.8 \ }

Total volume system = 35.0 ml + 17.5 ml = 52.5 ml.

\boxed{ \ [CH_3COOH] = \frac{2.625 \ mmol}{52.5 \ ml} = 0.05 \ M \ }

\boxed{ \ [CH_3COO^-] = \frac{2.625 \ mmol}{52.5 \ ml} = 0.05 \ M \ }

Step-4

And now it's time we calculate the pH value of this buffer system.

\boxed{ \ pH = pK_a + log\frac{[CH_3COO^-]}{[CH_3COOH]} \ }

\boxed{ \ pH = 5 - log \ 1.8 + log\frac{[0.05]}{[0.05]} \ }

\boxed{ \ pH = 5 - log \ 1.8 + log \ 1 \ }

\boxed{ \ pH = 5 - log \ 1.8 + 0 \ }

Thus, we get the pH value of this buffer system which is equal to 5 - log 1.8 or 4.74.

<h3>Learn more</h3>
  1. What is the pH of this buffer brainly.com/question/11437567
  2. Calculate the percent ionization (α) of formic acid solutions having the following concentrations (M). brainly.com/question/12198017  
  3. What is [S²⁻] at equilibrium? brainly.com/question/12593134
You might be interested in
The diagram below shows the different phase transitions that occur in matter. Three bars are shown labeled Solid, Liquid, and Ga
Grace [21]

The arrow that will represent the phase change that involves the same amount of energy as arrow 1 will be arrow 4.

<h3>Phase change</h3>

Arrow 1 represents a phase change from liquid to gas while arrow 4 represents a phase change from gas to liquid.

In other words, arrow 1 and arrow 4 are direct opposites of one another,

This means that if X amount of energy is required for arrow 1, the same amount of energy will be needed for arrow 4 but in the reverse direction.

More on phase change can be found here: brainly.com/question/12390797

#SPJ1

5 0
1 year ago
Hermit crabs and anemones have a mutualistic relationship. Which of these statements best describes how they interact?
liubo4ka [24]

Answer:  anemones cannot move very quickly, but they can sting predators. When an anemone is riding on a hermit crab's shell, the anemone protects the crab from predators.

Explanation:

5 0
2 years ago
What is the correct answer?
Korolek [52]

I think it's Almond Soy Milk because they're recommending your body's pH to be at 7.5 and the Almond Soy Milk is the answer with the closest pH to 7.5

3 0
3 years ago
Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0
2 years ago
What is the most common isotope of lithium?
Advocard [28]
Lithium-7 is the most common isotope of lithium.
7 0
3 years ago
Other questions:
  • Stronitum-90 has a half-life of 29 years. if a site held 2000 kg of this isotope, approximately what mass of strontium-90 would
    12·2 answers
  • A 10 gram sample of material, a, was placed in water. four grams of it, b, did not dissolve, even after being separated and plac
    15·1 answer
  • If the total mass of the products of a reaction is 90 gram, what was the total mass of the reactants
    13·1 answer
  • Which statement is true about the properties of the majority of the elements in the far left column of the periodic table?
    7·2 answers
  • _________ refers to the total energy of a system.
    5·2 answers
  • if 115 G of a substance reacts with 84 grams of another substance what will be the mass of the products after the reaction​
    12·1 answer
  • Identify two ways that the rate of a chemical reaction could be increased
    14·1 answer
  • If you were to observe the liquid evaporating away what would you see forming at
    15·1 answer
  • For the oxidation of ammonia
    14·1 answer
  • What kind of weather is associated with high air pressure?
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!