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lana66690 [7]
2 years ago
7

If one adds 0.1 mol of the weak acid hf (pk_a = 3.2) to a solution with a ph = 2, which species would be most abundant and how m

uch more abundant under such conditions?
Chemistry
1 answer:
zavuch27 [327]2 years ago
4 0

F^{-} is most abundant and 6310 times more than HF.

<h3>What is a strong and weak acid?</h3>

When an acid is dissolved in water, all of its molecules disintegrate, making the acid powerful.

When an acid is dissolved in water, only a small number of its molecules disintegrate, making the acid weak. Strong acids have a lower pH than weak acids.

The powerful acids include perchloric acid, chloric acid, nitric acid, sulfuric acid, hydrobromic acid, and hydroiodic acid.

Given:

Pka=3..2

pH=7

Let the volume be 1 liter

[HF]=01 M

pH=pka+log \frac{F^{-}}{HF} \\\\7=3.2+log\frac{F^{-}}{HF} \\3.8=log\frac{F^{-}}{HF}\\ \frac{F^{-}}{0.1}=10^{3.8} \\F^{-}=630.95 M

Now,

\frac{F^{-}}{HF}=\frac{630.95}{0.1}\\ =6309.57

F-:HF= 6309.57:1

Therefore, the most abundant is F^{-}and has 6310 times more than HF is F^{-}.

To know more about strong and weak acids, visit: brainly.com/question/12811944

#SPJ4

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Answer:

C

Explanation:

A negative deltaH means that the reaction has to give up heat in order to happen. You have to treat deltaH as a reactant. So the question is do you need to add heat to the reactants to make the products. If you do, deltaH is plus.

Heat is required to make a solid go to a gas. deltaH is plus. A is not the answer.

A lot of heat is required for B (something like 400 Kj / mole. Like A, deltaH is plus and B is not the answer.

C: The liquid has to give up heat in order for the this reaction to take place. C is the answer.

D requires heat. It is not the answer.

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What is the answer and why
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Read 2 more answers
Which of the following has the strongest buffering capacity? A. H2O B. 0.1 M HCl C. 0.1 M carbonic/bicarbonate (H2CO3/HCO3-) at
enyata [817]

Explanation:

(A)   As we know that carbonic acid (H_{2}CO_{3}) and Sodium bicarbonate (NaHCO_{3}) forms an acidic buffer.

Therefore, pH of an acidic buffer is given by Hendeerson-Hasselbalch equation as follows.

               pH = pK_{a} + log(\frac{[Salt]}{[Acid]}) ........... (1)

So mathematically,  if [Salt] = [Acid]  then \frac{[Salt]}{[Acid]} = 1 .

And,  log (\frac{[Salt]}{[Acid]}) = 0

Therefore, equation (1) gives us the following.

         pH = pK_{a} (when acid and salt are equal in concentration)

Hence, pK_{a} of H_{2}CO_{3} (carbonic acid) is 6.35.

And, with this we have following results.

In (A) and (D) we have the case \frac{[NaHCO_{3}]}{[H_{2}CO_{3}]}[/tex] i.e. [Salt] = [Acid].

Hence, for the cases pH = pK_{a} = 6.35.

(B)    [NaHCO_{3}] = 0.045 M and,  [H_{2}CO_{3}] = 0.45 M

Hence,   pH = 6.35 + log([NaHCO_{3}][[H_{2}CO_{3}])

                     = 6.35 + log(\frac{0.045}{0.45})

                     = 6.35 + (-1)

                     = 5.35

Therefore, it means that this buffer will be most suitable buffer as it has pH on acidic side and addition of slight excess base will not affect much of its pH value.

(C)    [NaHCO_{3}] = 0.45 M [H_{2}CO_{3}]

                          = 0.045 M

So,       pH = 6.35 + log(\frac{[NaHCO_{3}]}{[H_{2}CO_{3}]})

                  = 6.35 + log(\frac{0.45}{0.045})

                  = 6.35 + (+1)

                 = 7.35

This means that pH on Basic side makes it no more acidic buffer.

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