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3 years ago

A solution has a22.0*10^-6 H30^+ Calculate the OH^- and pH of this solution, be sure to show work.

Chemistry

1 answer:

mrs_skeptik [129]3 years ago

6 0

Answer:

pH= 4.66

OH^- = 4.55*10^-10

Explanation:

So H3O^+ means the same thing as [H^+].

so pH= -log(22.0*10^-6)

pH= 4.66

To find [OH^-]

[H^+] [OH^-] = 1*10^-14

1*10^-14 / 22.0*10^-6 = 4.55*10^-10= OH^-

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3 years ago

g Suppose a 49. L reaction vessel is filled with 1.5 mol of NO. What can you say about the composition of the mixture in the ves

irakobra [83]

The question is incomplete, the complete question is:

At a certain temperature, the equilibrium constant $K_{eq}$ for the following reaction is 0.74:

$NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)$

Suppose a 49.0 L reaction vessel is filled with 1.5 mol of $NO_3$ and 1.5 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium?

A. There will be very little $NO_3$ and NO.

B. There will be very little $NO_2$

C. Neither of the above is true.

Answer: The correct option is B. there will be very little $NO_2$

Explanation:

We are given:

Initial moles of $NO_3$ = 1.5 moles

Initial moles of NO = 1.5 moles

Volume of vessel = 49 L

As, moles of reactants and moles of products are equal, the volume term will not appear in the equilibrium constant expression.

Equilibrium constant for the reaction = 0.74

For the given chemical equation:

$NO_3(g)+NO(g)\rightleftharpoons 2NO_2(g)$

Initial: 1.5 1.5 -

At eqllm: 1.5-x 1.5-x 2x

The expression of equilibrium constant for the above reaction:

$K_{eq}=\frac{[NO_2]^2}{[NO_3]\times [NO]}$

Putting values in above equation, we get:

$0.74=\frac{(2x)^2}{(1.5-x)(1.5-x)}\\\\x=-1.13, 0.451$

Neglecting the negative value of equilibrium constant because concentration cannot be negative

Equilibrium concentration of $NO_3$ = (1.5 - x) = (1.5 - 0.451) = 1.049 moles

Equilibrium concentration of NO = (1.5 - x) = (1.5 - 0.451) = 1.049 moles

Equilibrium concentration of $NO_2$ = 2x = $(2\times 0.451)=0.902mol$

There are 3 possibilities:

If $K_{eq}$ , the reaction is reactant favored
If $K_{eq}>1$ , the reaction is product favored.
If $K_{eq}=1$ , the reaction is in equilibrium.

Here, the value of $K_{eq}=0.74$ , which is less than 1, therefore the reaction is reactant favored and we can say that there will be very little $NO_2$ .