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2 years ago

What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl

Chemistry

1 answer:

Norma-Jean [14]2 years ago

8 0

Answer:

$pH=4.63$

Explanation:

Hello!

In this case, since the ionization of ammonia, which is a weak base, is written as:

$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$

We can see that the ammonium ion is the conjugate acid whereas the hydroxide ions the conjugate base; that is why we use the Henderson-Hasselbach equation to compute the pH, given the pKb of ammonia 4.75:

$pH=pKb+log(\frac{[conj\ acid]}{[base]} )$

In such a way, for the given moles of ammonia, base, and those of ammonium chloride, conjugate acid form, we obtain:

$pH=4.75+log(\frac{0.15}{0.2} )\\\\pH=4.63$

Best regards!

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Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

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Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

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What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl​

What's the pH of a solution composed of 0.2 moles of NH3 and 0.15 moles of nh4cl