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Aleks04 [339]
3 years ago
11

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) Pb+2(aq) + 2Cl-(aq).

Chemistry
1 answer:
drek231 [11]3 years ago
7 0

Answer:

Option A and Option B

Explanation:

The concentrations of Pb+2(aq) and Cl-(aq) are expected to decrease (because you have initially very high values of [Pb2+] and [Cl-], they should decrease so that you equilibrium is reestablished, so that Qsp = Ksp).

Solids aren't measured using concentrations but the AMOUNT of PbCl2(s) is expected to increase.

It IS possible to have a system with Ksp > 1.8 x 10^-2 -- if you increase the temperature, then solubility and Ksp are both increased. It is also possible to have a system with Qsp > Ksp (as is true in this case). However, if Qsp > Ksp, the system will shift so that reequilibrium is reestablished. Over time, Qsp will be come equal to Ksp.

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You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution when ex
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a. 1.78x10⁻³ = Ka

2.75 = pKa

b. It is irrelevant.

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a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.

Equilibrium is:

HA ⇄ H⁺ + A⁻

And Ka is defined as:

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The HA reacts with the base, XOH, thus:

HA + XOH → H₂O + A⁻ + X⁺

As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻

That means:

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It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:

pH = pKa + log₁₀ [A⁻] / [HA]

Replacing:

2.75 = pKa + log₁₀ [A⁻] / [HA]

As [HA] = [A⁻]

2.75 = pKa + log₁₀ 1

<h3>2.75 = pKa</h3>

Knowing pKa = -log Ka

2.75 = -log Ka

10^-2.75 = Ka

<h3>1.78x10⁻³ = Ka</h3>

b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.

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