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3 years ago

What is the solubility of NH4Cl at 50°C?

Chemistry

2 answers:

astra-53 [7]3 years ago

8 0

The correct answer is E. 50 grams

The solubility of Ammonium Chloride has an increasing trend as the temperature increases. Experiments show that the solubility at 323K is approximately 50.4 grams per 100 grams o water thus this value is close to choice D, 50 grams

maw [93]3 years ago

3 0

<u>Answer:</u> The correct answer is Option E.

<u>Explanation:</u>

Solubility is defined as the amount of substance (known as solute) which is dissolved in a unit volume of liquid substance (known as solvent) to produce a saturated solution at specific temperature and pressure. It is expressed in number of moles of solute present per 100 grams of solvent.

This property is directly dependent on temperature. As, the temperature increases, the solubility of the solute increases and vice-versa.

$\text{Solubility}\propto \text{Temperature}$

$NH_4Cl$ is an ionic salt whose solubility increases as we increase the temperature.

The solubility of ammonium chloride at 50°C is 50 grams.

Hence, the correct answer is Option E.

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Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for

Katyanochek1 [597]

Answer:

See explanation below

Explanation:

There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.

However we do not have a pH value here.

The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.

Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:

LiOH ---> Strong. If you try to dissociate :

LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.

HF -----> Weak

HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.

HCl -----> Strong

This is actually one of the strongest acid.

NH₃ ------> Weak

Now writting the Ka and Kb expressions:

Ka = [H⁺] [F⁻] / [HF]

Kb = [NH₄⁺] [OH⁻] / [NH₃]

Finally, to calculate the [OH⁻] we need to use the following expression:

Kw = [H⁻] [OH⁻]

Solving for [OH⁻] we have:

[OH⁻] = Kw / [H⁺]

Remember that the value of Kw is 1x10⁻¹⁴. So replacing:

[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶

[OH⁻] = 1.43x10⁻⁹ M

And now, multiplying by 10¹⁰ we have:

[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰

Hope this helps

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I hope it helps, Regards.

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