All categories

4 years ago

If the kb of a weak base is 3.3 Ã 10-6, what is the ph of a 0.17 m solution of this base?

Chemistry

1 answer:

Nana76 [90]4 years ago

8 0

The relation between base equilibrium constant and degree of dissociation is

C x^2 = Kb

x is degree of dissociation

C x^2 = 3.3 * 10^-6

(0.17) x^2 = 3.3 * 10^-6

x = 4.24 * 10^-3

Concentration of hydroxyl ion is

[OH-] = c x = 0.17 * 4.24 * 10^-3 = 0.72 * 10^-3 M

PoH = - log [OH-] = 2.85

PH = 14 - pOH = 11.15

Hence, pH of solution is 11.15

You might be interested in

Help please i give brainlist

shtirl [24]

The order of the answers are as follows:
B
C
D
A

5 0

4 years ago

Pick the correct model

frutty [35]

Answer:

A and B is the answer as there structure change

3 0

3 years ago

The diagram shows the structure of an animal cell.

Vera_Pavlovna [14]

Answer:

Explanation:

The cell membrane is the protective, semi-permeable membrane surrounding a cell and its contents.

8 0

3 years ago

Read 2 more answers

Which of the following represents the smallest mass? I.) 23 cg II.) 2.3 x 10-3 μg III.) 0.23 mg IV.) 0.23 g V.) 2.3 x 10-2 kg

jenyasd209 [6]

The smallest mass among the choices is Choice (ii):2.3 x 10-3 μg

This is so because, the prefix μ signifies a factor of 10^-6.

For choice I:

23 cg means 23 centigrams

In essence = 0.23grams

For choice II:

2.3 × 10-3 μg means 2.3 × 10-9 grams

For choice III:

0.23 mg means 0.23 × 10-3 grams

For choice IV:

0.23 grams simply means 0.23grams

For choice V:

2.3 × 10-2 kg means 23 grams

Ultimately, Choice II represents the smallest mass.

Read more;

brainly.com/question/17192728

7 0

2 years ago

Gaseous methane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water. If 2.59 g of water is produc

max2010maxim [7]

Answer: The percent yield of the water is 31.98 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For methane:

Given mass of methane = 6.58 g

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane}=\frac{6.58g}{16g/mol}=0.411mol$

For oxygen gas:

Given mass of oxygen gas = 14.4 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

$\text{Moles of oxygen gas}=\frac{14.4g}{32g/mol}=0.45mol$

The chemical equation for the combustion of methane is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

By Stoichiometry of the reaction:

2 moles of oxygen gas reacts with 1 mole of methane

So, 0.45 moles of oxygen gas will react with = $\frac{1}{2}\times 0.45=0.225mol$ of methane

As, given amount of methane is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction

2 moles of oxygen gas produces 2 moles of water

So, 0.45 moles of oxygen gas will produce = $\frac{2}{2}\times 0.45=0.45$ moles of water

Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 0.45 moles

Putting values in equation 1, we get:

$0.45mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.45mol\times 18g/mol)=8.1g$

To calculate the percentage yield of water, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of water = 2.59 g

Theoretical yield of water = 8.1 g

Putting values in above equation, we get:

$\%\text{ yield of water}=\frac{2.59g}{8.1g}\times 100\\\\\% \text{yield of water}=31.98\%$

Hence, the percent yield of the water is 31.98 %

4 0

3 years ago