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3 years ago

6

An aqueous solution is prepared by dissolving the salt formed by the neutralization of a weak acid by a weak base. Which stateme

nt about the solution is correct?
a. the solution is strongly basicb. the solution is weakly basicc. the solution is neutrald. the solution is acidice. the values for Ka and Kb for the species in solution must be known before a prediction can be made

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Answer: The values for Ka and Kb for the species in solution must be known before a prediction can be made

Explanation:

1. If Ka is greater than Kb, the salt solution is acidic

2. If Ka is lesser than Kb, the salt solution is basic

3. If Ka is equal to Kb, the salt solution is acidic

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An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

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The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=6.798 Torr$

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Explanation:

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=P_1=13.2 Torr$

Partial pressure of the $Cl_2=P_2=13.2 Torr$

Partial pressure of the $PCl_5=P_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{P_1}{P_1\times P_2}$

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At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=P_1'=13.2 Torr$

Partial pressure of the $Cl_2=P_2'=?$

Partial pressure of the $PCl_5=P_3'=217.0 Torr$

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$P=P_1'+P_2'+P_3'$

$263.0Torr=13.2 Torr+P_2'+217.0 Torr$

$P_2'=32.8 Torr$

$PCl_3(g) + Cl_2(g)\rightleftharpoons PCl_5(g)$

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{P_3'}{P_1'\times P_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

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