Which of the following correctly describes interactions between Earth's spheres?

A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f

Svetllana [295]

Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid, CH 3 COOH , a weak acid, and sodium hydroxide, NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not. If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal 7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over 7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq] + OH − (aq] → CH 3 COO − (aq] + H 2 O (l]

Notice that:

1 mole of acetic acid will react with: 1 mole of sodium hydroxide, shown here as hydroxide anions, OH − , to produce 1 mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

c = n / V ⇒ n = c ⋅ V

n acetic = 0.20 M ⋅ 25.00 ⋅ 10 − 3 L = 0.0050 moles CH3 COOH

and

n hydroxide = 0.10 M ⋅ 40.00 ⋅ 10 − 3 L = 0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

n acetic remaining = 0.0050 − 0.0040 = 0.0010 moles

The reaction will also produce 0.0040 moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH = p K a + log ( [ conjugate base ] / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total = V acetic + V hydroxide

V total = 25.00 mL + 40.00 mL = 65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ] = 0.0010 moles / 65.00 ⋅ 10 − 3 L = 0.015385 M

and

[ CH 3 COO − ] = 0.0040 moles / 65 ⋅ 10 − 3 L = 0.061538 M

The p K a of acetic acid is equal to 4.75

Thus the pH of the solution will be:

pH = 4.75 + log ( 0.061538 M / 0.015385 M )

pH = 5.35

5 0

3 years ago

Read 2 more answers

Select the correct term to complete each sentence.

leva [86]

Answer: 1. halve

2. halve

3. double

Explanation:

The relationship between wavelength and energy of the wave follows the equation:

$E=\frac{hc}{\lambda}$

E= energy

$\lambda$ = wavelength of the wave

h = Planck's constant

c = speed of light

Thus as wavelength and energy have inverse realation, when wavelength will halve , energy will double.

2. The between wavenumber and energy of the wave follows the equation:

$E=h\times c\times \bar{\nu}$

E= energy

$\bar{\nu}$ $\nubar$ = wavenumber of the wave

h = Planck's constant

c = speed of light

Thus as wavenumber and energy have direct relation, when wavenumber will halve , energy will be halved.

3. The relationship between energy and frequency of the wave follows the equation:

$E=h\times \nu$

where

E = energy

h = Planck's constant

$\nu$ = frequency of the wave

Thus as frequency and energy have direct realation, when frequency will double , energy will double.

3 0

3 years ago

What are formed when sodium ions and chlorine ions combine to produce nacl?

GREYUIT [131]

<span>bright yellow light and lots of heat-energy.</span>

4 0

3 years ago

For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →

Vinvika [58]

Answer : The oxidizing element is N and reducing element is O.

$KNO_{3}$ is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

$KNO_{3}$ act as an oxidizing agent.

The oxidation number of N in $KNO_{3}$ is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in $KNO_{2}$ is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number reduced this means $KNO_{3}$ itself get reduced to $KNO_{2}$ and it can act as an oxidizing agent.

$KNO_{3}$ act as a reducing agent.

$KNO_{3} \rightarrow KNO_{2} +O_{2}$

The oxidation number of O in $KNO_{3}$ is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in $O_{2}$ is Zero (o).

Now, we conclude that the oxidation number increases this means $KNO_{3}$ itself get oxidized to $O_{2}$ and it can act as reducing agent.

4 0

4 years ago

Read 2 more answers

Which section of the reaction represents the reactants

Diano4ka-milaya [45]

A should be the products and D should be the reactants. So D should be the answer.

8 0

3 years ago