All categories

3 years ago

The pH of an acetic acid solution is 4.0. (10n= 1en).

Chemistry

1 answer:

frosja888 [35]3 years ago

8 0

Answer:

1. [H+] = 1 x 10^-4mol/L

2. [OH-] = 1 x 10^-10mol/L

Explanation:

1. From the question given, the pH is 4.0. We can obtain the value of the value of the hydrogen ion concentration [H+] as follows:

pH = —Log[H+]

4 = —Log[H+]

—4 = Log[H+]

Anti-log of —4

[H+] = 1 x 10^-4mol/L

2. Recall: [H+] x [OH-] = 1 x 10^-14

But [H+] = 1 x 10^-4mol/L

1 x 10^-4 x [OH-] = 1 x 10^-14

Divide both side by 1 x 10^-4mol/L

[OH-] = 1 x 10^-14/1 x 10^-4

[OH-] = 1 x 10^-10mol/L

You might be interested in

In the equation:

Sliva [168]

3 MgCl₂ + 1 Na₃P = 1 Mg₃P₂ + 6 NaCl

Mg₃Cl₂ = Magnesium Phosphide

NaCl = Sodium chloride

hope this helps!

6 0

3 years ago

Read 2 more answers

Explain how growth in different from development

lubasha [3.4K]

In growth you can see a change. In developing it takes time.

3 0

4 years ago

Why does increasing the temperature of two reactants in solution make a

gogolik [260]

The two molecules will only react if they have enough energy. By heating the mixture, you are raising the energy levels of the molecules involved in the reaction. Increasing temperature also means the molecules are moving around faster and will therefore "bump" into each other more often.

5 0

3 years ago

Read 2 more answers

What is the pH of an aqueous solution of 1 M CH3COOH (pKa=4)

Mrac [35]

Answer:

pH = 2.

Explanation:

A weak acid is in equilibrium with its ions in a solution, so it must have an equilibrium constant (Ka). And, pKa = -logKa

$Ka = 10^{-pKa}$

Ka = 10⁻⁴

So, for CH₃COOH the equilibrium must be:

CH₃COOH(aq) ⇄ H⁺(aq) + CH₃COO⁻(aq)

1 M 0 0 Initial

-x +x +x Reacted

1-x x x Equilibrium

And the equilibrium constant:

$Ka = \frac{[H+]x[CH3COO-]}{[CH3COOH]}$

$10^{-4} = \frac{x^2}{1-x}$

Supposing x << 1:

10⁻⁴ = x²

x = √10⁻⁴

x = 10⁻² M, so the supposing is correct.

So,

pH = -log[H⁺]

pH = -log10⁻²

pH = 2

8 0

4 years ago

Carbon disulfide burns in oxygen to yield carbon dioxide and sulfur dioxide according to the following chemical equation. CS2(l)

Semmy [17]

Answer:

a) the limiting reactant is 02

b) There will remain 0.667 moles of CS2

c) There will be formed 0.333 moles oof CO2 and 0.667 moles of SO2

Explanation:

Step 1: Data given

Number of moles of CS2 = 1.00 mol

Number of moles of O2 = 1.00 mol

Molar mass of O2 = 32 g/mol

Molar mass of CS2 = 76.14 g/mol

Step 2: The balanced equation

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

Step 3: Calculate the limiting reactant

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

O2 is the limiting reactant. It will completely be consumed.(1.00 mol).

CS2 is in excess. There will react 1.00/ 3 = 0.333 moles

There will remain 1.00 - 0.333 = 0.667 moles of CS2

Step 4: Calculate moles of CO2 and SO2

For 1 mole of CS2 we need 3 moles of O2 to produce 1 mol of CO2 and 2 moles of SO2

For 1.00 mol of O2 we have 1.00/3 = 0.333 moles CO2

For 1.00 mol of O2 we have 1.00 /(3/2) = 0.667 moles of SO2

8 0

3 years ago