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babunello [35]
3 years ago
5

Examining the equations or equilibrium constants related to a base, salt, or an acid is an indirect way to determine strength of

an electrolyte. The strength of an electrolyte can be examined directly by placing a solution into a circuit so that the voltage or amount of current can be measured. Although conductivity refers to the flow of charged species, we usually examine conductivity with respect to resistance. As the name implies, a solution that does not conduct electricity very well also has a very high resistance. If solutions containing various acids, bases, and salts were prepared and connected to a circuit that powers a light bulb, the strength of the electrolyte could be estimated by examining the intensity of the light bulb. Complete the following sentences regarding the experimental determination of the electrolyte strength for various molecules dissolved in solution.
a. strong electrolyte
b. weak electrolyte
c. non-electrolyte
1. A 20 mL solution containing 2 mmol of Ca3(PO4)2(s) was integrated into to a circuit that powers a light but, When the power supply was turned on, the light bulb produced a glow. Ca3(PO4)2(s) is__.
2. A 20 mL, solution containing 2 mmol of C12H22O11(aq) was integrated into a circuit that powers a light bulb. When the power supply was turned on the light bulb remained off. C12H22O11(aq) is___.
3. A 20 mL, solution containing 2 mmol of HF(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb faintly flickered. HF(g) is an____.
4. A 20 mL solution containing 2 mmol of LiOH(g) was integrated into to a circuit that power a light bulb. When the power supply was turned on the light bulb produced a bright glow LiOH(s) is a____.
Chemistry
1 answer:
Gwar [14]3 years ago
7 0

Answer:

1 - Weak electrolyte

2- Non electrolyte

3- Weak electrolyte

4- Strong electrolyte

Explanation:

A strong electrolyte refers to an electrolyte that decomposes completely in solution. This means that there are more charge carriers in solution when a strong electrolyte is dissolved in water. A strong electrolyte produces a strong glow. LiOH is a strong electrolyte.

A weak electrolyte is not completely dissociated in water. Only a small amount dissociates in water. HF is a weak electrolyte. A weak electrolyte does not produce a bright light.

A non-electrolyte does not dissociate in solution at all hence it does not power a bulb E.g  C12H22O11.

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
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Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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atomic mass of Thallium <span>204.41 u
</span>
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