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Nimfa-mama [501]
3 years ago
15

1. Determine whether the following hydroxide ion concentrations ([OH−]) correspond to acidic, basic, or neutral solutions by est

imating their corresponding hydronium ion concentrations ([H3O+] using the ion product constant of water (Kw).
Kw = [H3O+][OH−] = [1×10−7 M][1×10−7 M] = 1×10−14 M
Hydronium ion concentration [H3O+] Solution condition
Greater than 1×10−7 M Acidic
Equal to 1×10−7 M Neutral
Less than 1×10−7 M Basic
Drag the appropriate items to their respective bins.
1. [OH−] = 6×10−12 M
2. [OH−] = 9×10−9 M
3. [OH−] = 8×10−10 M
4. [OH−] = 7×10−13 M
5. [OH−] = 2×10−2 M
6. [OH−] = 9×10−4 M
7. [OH−] = 5×10−5 M
8. [OH−] = 1×10−7 M
A. Acidic
B. Neutral
C. Basic
2. A solution has [H3O+] = 5.2×10−5M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [OH−] of the solution.
3. A solution has [OH−] = 2.7×10−2M . Use the ion product constant of water
Kw=[H3O+][OH−]
to find the [H3O+] of the solution.
Chemistry
1 answer:
djyliett [7]3 years ago
3 0

Answer:

Question 1.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

[H₃O⁺] = 3.70 * 10⁻¹¹ M

Explanation:

The ion product constant of water  Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M² is a constant which gives the product of the concentrations of hydronium and hydroxide ions of dissociated pure water. The concentrations of the two ions are both equal to 1 * 10⁻⁷ in pure water.

A solution that has [OH⁻] greater than 1 * 10⁻⁷ is basic while one with [OH⁻] less than 1 * 10⁻⁷ is acidic.

1. [OH−] = 6×10−12 M  is less than 1 * 10⁻⁷, therefore is acidic.

2. [OH−] = 9×10−9 M  is less than 1 * 10⁻⁷, therefore is acidic.

3. [OH−] = 8×10−10 M  is less than 1 * 10⁻⁷, therefore is acidic.

4. [OH−] = 7×10−13 M  is less than 1 * 10⁻⁷, therefore is acidic.

5. [OH−] = 2×10−2 M  is greater than 1 * 10⁻⁷, therefore is basic.

6. [OH−] = 9×10−4 M  is greater than 1 * 10⁻⁷, therefore is basic.

7. [OH−] = 5×10−5 M  is greater than 1 * 10⁻⁷, therefore is basic.

8. [OH−] = 1×10−7 M  is equal to 1 * 10⁻⁷, therefore is neutral

Question 2:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[OH⁻] = 1 * 10⁻¹⁴ M²/ [H₃O⁺]

[OH⁻] = 1 * 10⁻¹⁴ M²/5.2*10⁻⁵ M

[OH⁻] = 1.92 * 10⁻⁸ M

Question 3:

Kw = [H₃O⁺][OH⁻] = 1 * 10⁻¹⁴ M²

[H₃O⁺][OH⁻] = 1 * 10⁻¹⁴

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ [OH⁻]

[H₃O⁺] = 1 * 10⁻¹⁴ M²/ 2.7 * 10⁻² M

[H₃O⁺] = 3.70 * 10⁻¹¹ M

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Answer:- B: 1.10g H_2 is the right answer.

Solution:- The balanced equation is:

O_2(g)+2H_2(g)\rightarrow 2H_2O(g)

We have been given with 8.75 grams of oxygen and asked to calculate the grams of hydrogen needed to react with given grams of oxygen according to the balanced equation.

From balanced equation, 1 mole of oxygen reacts with 2 moles of hydrogen.

So, let's convert grams of oxygen to moles and multiply it by the mole ratio to calculate the moles of hydrogen that are easily converted to grams on multiplying by it's molar mass.

The complete set up looks as:

8.75g O_2(\frac{1mole}{32g})(\frac{2mole H_2}{1mole O_2})(\frac{2.02g}{1mole})

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In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +
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<u>Answer:</u> The pH of the solution is 1.136

<u>Explanation:</u>

To calculate the moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

  • <u>For ammonia:</u>

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol

  • <u>For nitric acid:</u>

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

                       NH_3+HNO_3\rightarrow NH_4NO_3

At t=0             0.0178   0.022

Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

pH=-\log[H^+]

where,

[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M

Putting values in above equation, we get:

pH=-\log(0.0731)\\\\pH=1.136

Hence, the pH of the solution is 1.136

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3 years ago
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