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Ann [662]
4 years ago
15

Which of the following is true of a buffered solution? Group of answer choices All of these are true The solution resists any ch

ange in its [H+]. The pH of the solution will not change much even if a concentrated acid is added. Any H+ ions added will react with a conjugate base of a weak acid already in thye solution. The pH of the solution will not change much even if a strong base is added.
Chemistry
1 answer:
olya-2409 [2.1K]4 years ago
5 0

Answer:

All of these are true

Explanation:

A buffer solution in chemistry is a solution that resists changes in pH when an acid or base is added to it. It is a solution that contains a weak acid and its conjugate base (anion) or a weak base and its conjugate acid.

A buffer is able to resist a change in pH due to the conjugate base and conjugate acid of the weak acids and bases contained in them respectively. The conjugate base/acid are present in an equilibrium quantity with their acid/base counterparts and help to neutralize or react with any additional H+ or OH- from an acid or base added to their solution.

However, when a strong acid or base is added to the buffer solution, there is only a slight change which practically does not change the pH of the solution.

Hence, all of the above options about a buffered solution is true.

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For the reaction POCl3(g) ⇀↽ POCl(g) + Cl2(g) Kc = 0.30. An initial 0.3 moles of POCl3 are placed in a 3.1 L container with init
olya-2409 [2.1K]

<u>Answer:</u> The final concentration of POCl_3 is 0.019767 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of the solution}}

Initial moles of POCl_3 = 0.3 moles

Volume of solution = 3.1 L

\text{Initial concentration of }POCl_3=\frac{0.3}{3.1}=0.097M

For the given chemical equation:

                        POCl_3(g)\rightleftharpoons POCl(g)+Cl_2(g)

<u>Initial:</u>                0.097

<u>At eqllm:</u>         0.097-x             x              x

The expression of K_c for above equation follows:

K_c=\frac{[POCl][Cl_2]}{[POCl_3]}

We are given:

K_c=0.30

Putting values in above expression, we get:

0.30=\frac{x\times x}{(0.097-x)}\\\\x=0.077233,-0.377

Neglecting the negative value of 'x' because concentration cannot be negative.

So, final concentration of POCl_3 = (0.097 - x) = (0.097 - 0.077233) = 0.019767 M

Hence, the final concentration of POCl_3 is 0.019767 M

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Explanation:

Using integrated rate law for first order kinetics as:

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[A_0] is the initial concentration

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