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Maksim231197 [3]
2 years ago
7

Hydrobromic acid, HBr, is a strong acid. Hydrofluoric acid, HF, is a weak acid. If you prepared equal

Chemistry
1 answer:
snow_tiger [21]2 years ago
4 0
HBr and HF are both monoprotic Arrhenius acids—that is, in aqueous solution, they dissociate and ionize to give hydrogen ions. A strong acid ionizes completely; a weak acid ionizes partially.

In this case, HBr, being a strong acid, would ionize completely in water to yield H+ and Br- ions. However, HF, being a weak acid, would ionize only to a limited extent: some of the HF molecules will ionize into H+ and F- ions, but most of the HF will remain undissociated.

pH is, by definition, a measurement of the concentration of hydrogen ions in solution (pH = -log[H+]). A higher concentration of hydrogen ions gives a lower pH, while a lower concentration of hydrogen ions gives a higher pH. At 25 °C, a pH of 7 indicates a neutral solution; a pH less than 7 indicates an acidic solution; and a pH greater than 7 indicates a basic solution.

If we have equal concentrations of HBr and HF, then the HBr solution will have a greater concentration of hydrogen ions in solution than the HF solution. Consequently, the pH of the HBr solution will be less than the pH of the HF solution.

Choice A is incorrect: Strong acids like HBr dissociate completely, not partially.

Choice B is incorrect: While the initial concentration of HBr and HF are the same, the H+ concentration in the HBr solution is greater. Since pH is a function of H+ concentration, the pH of the two solutions cannot be the same.

Choice C is correct: A greater H+ concentration gives a lower pH value. The HBr solution has the greater H+ concentration. Thus, the pH of the HBr solution would be less than that of the HF solution.

Choice D is incorrect for the reason why choice C is correct.

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Why does the flame go out on the Bunsen burner when the gas is turned off? (use the collision theory in your answer)
Gre4nikov [31]

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Explanation:

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18. give the product for the following reactions sequence. a) 3-pentanone b) 2-pentanone c) propanone d) 2-butanone e) methyl pr
serg [7]

The product for the following reaction are :

  • 2 - butanone
  • CH₃CH₂COCl
  • CH₃CH₂CONHCH₃

T carboxylic acid is an organic compound. the functional group of carboxylic acid is carboxy. general formula is given as : R - COO⁻.

The reactions are given as follows :

1) the reduction of carboxylic acid into ketone with the organometallic compound is given as follows :

CH₃CH₂COOH     +  CH₃Li   ------>   CH₃CH₂COCH₃

                                                                  2 butanone

2) the reaction of carboxylic acid with SOCl₂ is given as follows :

CH₃CH₂COOH    +   SOCl₂   ----->   CH₃CH₂COCl

3) the reaction of carboxylic acid with CH₃NH₂ is given as follows

CH₃CH₂COOH  +   CH₃NH₂   ----->   CH₃CH₂CONHCH₃

Thus,  The product for the following reaction are :

  • 2 - butanone
  • CH₃CH₂COCl
  • CH₃CH₂CONHCH₃

To learn more about carboxylic acid here

brainly.com/question/4721247

#SPJ4

5 0
1 year ago
[H3O+] [OH−] pH Acidic or Basic 3.5×10−3 _____ _____ _____ _____ 3.8×10−7 _____ _____ 1.8×10−9 _____ _____ _____ _____ _____ 7.1
ankoles [38]

Answer:

See explanation below

Explanation:

I'm assuming this is a table you need to complete, so, you are not putting this in order, but I already found it in another place. let's do a little summary of the expressions we need to use in order to complete the chart.

To calculate pH we need the following expression:

<em>pH = -log[H3O+] (1)</em>

From this expression we can solve for [H3O+] in case we need it:

<em>[H3O+] = 10^(-pH)   (2)</em>

When we have [OH-] we calculate the pOH and from there, the pH:

<em>pOH = -log[OH-]  (3)</em>

and [OH-]:

<em>[OH-] = 10^(-pOH) (4)</em>

Finally to get the pH from pOH:

14 = pH + pOH

<em>pH = 14 - pOH (5)</em>

With these 5 expressions we can complete the chart. In picture 1, you have the actual chart.

To know if it's acidic or basic, that depends on the value of pH.

If pH <7 it's acidic

If pH >7 it's basic

If pH = 7 it's neutral

<u><em>First case:</em></u>

[H3O+] = 3.5x10^-3

pH = -log(3.5x10^-3) = 2.46  (acidic)

pOH = 14 - 2.46 = 11.54

[OH-] = 10^(-11.54) = 2.88x10^-12 M

<u><em>Second case:</em></u>

pOH = -log(3.8x10^-7) = 6.42

pH = 14 - 6.42 = 7.58 (it's basic)

[H3O+] = 10^(-7.58) = 2.63x10^-8 M

<u><em>Third case:</em></u>

pH = -log(1.8x10^-9) = 8.74 (it's basic)

pOH = 14 - 8.74 = 5.26

[OH-] = 10^(-5.26) = 5.5x10^-6 M

<u><em>Fourth case:</em></u>

[H3O+] = 10^(-7.15) =7.08x10^-8 M   (Basic)

pOH = 14 - 7.15 = 6.85

[OH-] = 10^(-6.85) = 1.41x10^-7 M

Hope this can help you

7 0
3 years ago
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