Is sulfur dioxide a element?

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

What is the wavenumber of the radiation emitted when a hydrogen

LUCKY_DIMON [66]

Answer: Wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

Explanation:

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

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

where,

E = energy of the radiation = $1.634\times 10^{-18}J$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength of radiation = ?

Putting values in above equation, we get:

$1.634\times 10^{-18}J=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{\lambda}\\\\\lambda=12.16\times 10^{-8}m$

$\bar {\nu}=\frac{1}{\lambda}=\frac{1}{12.16\times 10^{-8}}=0.08\times 10^{8}m^{-1}$

Thus wavenumber of the radiation emitted is $0.08\times 10^{8}m^{-1}$

8 0

4 years ago

All chemical reactions have a conservation of(1) mass, only(2) mass and charge, only(3) charge and energy, only(4) mass, charge,

Semenov [28]

All chemical reactions have a conversation of mass and energy.

Because:- There are only two laws for conversation in a chemical reaction. The conversation of mass, no mass can be created nor destroyed. Also, the law of conversation of mass states that no energy can be created nor destroyed in a chemical reaction. The charge can obviously change because they can bond and change charges.

7 0

3 years ago

Read 2 more answers

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

412.072 g Cl₂ ≈ 412 g Cl₂

5 0

3 years ago

Name the 3 seperate metals group

umka21 [38]

Akali Metals

Akali-Earth Metals

and Other Metals

8 0

3 years ago