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Vikki [24]
4 years ago
14

According to one acid-base theory a molecule acts as an acid when the molecule

Chemistry
2 answers:
Elden [556K]4 years ago
7 0

Answer:idk tbh

Explanation:

Solnce55 [7]4 years ago
4 0
Donates an H+ ion {Arrhenius theory}
You might be interested in
A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L
Pavel [41]

Answer:

The pH does not decrease drastically because the HCl reacts with the <u>sodium azide (NaN₃)</u> present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l)  ⇄ N₃⁻(aq) + H₃O⁺(aq)          

The pH of this solution is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq)  ⇄  HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles

Now, the number of moles of HN₃ is:

\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles

Then, the pH of the buffer solution after the addition of HCl is:

pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.    

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

6 0
3 years ago
(a) Kw = 1.139 × 10⁻¹⁵ at 0°C and 5.474 × 10⁻¹⁴, find [H₃O⁺] and pH of water at 0°C and 50°C.
kondor19780726 [428]

The  value of [H₃O⁺] and  pH of water at 0°C and 50°C.

0°C value of [H₃O⁺] = 3.375 x 10⁻⁸

50°C value of [H₃O⁺]  = 2.340 x 10⁻⁷

pH of water at 0°C  =  pH = 7.4717

pH of water at 50°C  =  pH  = 6.6308

<h3>pH of water at different level:</h3>

Water with a pH less than 7 is considered acidic, while water with a pH greater than 7 is considered basic. The normal pH range for surface water systems is 6.5 to 8.5, and for groundwater systems is 6 to 8.5. Alkalinity is a measure of a water's ability to withstand a pH change that would cause it to become more acidic.

<h3>According to the given information:</h3>

0°C  =  Kw = 1.139 × 10⁻¹⁵

50°C = 5.474 × 10⁻¹⁴

Solving at  0°C  for  [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]

are equal [H₃O⁺] = [OH⁻].

                                                Kw  =  [H₃O⁺] [OH⁻]

                                                  Kw  =  [H₃O⁺]²

                                                  [H₃O⁺] =  √Kw

                                                            =  √1.139 × 10⁻¹⁵

                                                            = 3.375 x 10⁻⁸

                                                pH = -log[H₃O⁺]

                                                      = -log 3.375 x 10⁻⁸

                                                      = 7.4717

Solving at  50°C  for  [H₃O⁺] and pH. water is neutral therefore its[H₃O⁺] [OH⁻]

are equal [H₃O⁺] = [OH⁻].

                                                 Kw  =  [H₃O⁺] [OH⁻]

                                                  Kw  =  [H₃O⁺]²

                                                  [H₃O⁺] =  √Kw

                                                             = √ 5.474 × 10⁻¹⁴

                                                            = 2.340 x 10⁻⁷ M

                                                 pH = -log[H₃O⁺]

                                                       =  -log2.340 x 10⁻⁷

                                                   pH  = 6.6308

The  value of [H₃O⁺] and  pH of water at 0°C and 50°C.

0°C value of [H₃O⁺] = 3.375 x 10⁻⁸

50°C value of [H₃O⁺]  = 2.340 x 10⁻⁷

pH of water at 0°C  =  pH = 7.4717

pH of water at 50°C  =  pH  = 6.6308

To know more about pH of water visit:

brainly.com/question/13822050

#SPJ4

I understand that the question you are looking for is:

Kw = 1.139 × 10⁻¹⁵ at 0°C and 5.474 × 10⁻¹⁴, find [H₃O⁺] and pH of water at 0°C and 50°C. find the  [H₃O⁺] and  pH of water at 0°C and 50°C.

 

3 0
1 year ago
(HELP FAST)Which of the following always changes when a substance undergoes a<br> chemical change?
Lera25 [3.4K]
I’m pretty sure it’s a
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Hhow do you make slime? best answer might get a brainy i need it for a party ASAP
strojnjashka [21]
Here is how i made slime when i had a buisiness.

Add glue to a bowl.
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4 0
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Archy [21]

Answer:

Pressure

Explanation:

The variable that describes how often the particles in a sample of gas collides with each other and the walls of the container is the pressure.

The pressure of a gas is the combined force with which the gas molecules bombard a unit area of the wall of the container i.e the sum of all the tiny pushes on the unit area of the wall of the container.

The various units of gas pressure atmosphere, millimeters of Hg, torr, pascal, newton per meter squared.

  • Temperature is the degree of hotness or coldness of a body
  • Volume is the space the gases occupies.
3 0
3 years ago
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