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umka21 [38]
3 years ago
11

Which of the following statements is/are true? I: The conjugate acid of C6H5NH2 is C6H5NH3+ II: An acid with Ka = 1x10−3 is stro

nger than an acid with Ka = 1x10−4 III: A strong acid will be completely dissociated at equilibrium A) Only I B) Only II C) I and III D) II and III E) I, II, and III
Chemistry
1 answer:
Anna [14]3 years ago
5 0

Answer:

E. I, II, and III.

Explanation:

I. C6H5NH2 + H2O --> C6H5NH3^+ + OH-

C6H5NH2 is a weak base means it is a good proton donor (strong conjugate acid).

II. Any aqueous acid with a pKa value of less than 0 is almost completely deprotonated and is considered a strong acid.

pka1 = - log[1 x 10^-3]

= 2

pka1 = - log[1 x 10^-4]

= 3

pka1 is a stronger acid.

III. A strong acid will be completely dissociated at equilibrium.

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2 years ago
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
Morgarella [4.7K]

Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

Cu(OH)_2 -> Cu^{2+} + 2OH^-

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

moles Cu(OH)_2 = 4.96E-8 mol/L * 1 L = 4.96E-8 moles

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