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Dovator [93]
3 years ago
11

Which statement is true for a solution when its concentration of hydroxide ions increases? It becomes more acidic. It becomes le

ss alkaline. Its pH value increases. Its pH value decreases.
Chemistry
2 answers:
alexira [117]3 years ago
6 0

Answer: Option (c) is the correct answer.

Explanation:

When we increase the concentration of a substance then it means we are increasing number of molecules of the substance. So, when there is increase in concentration of hydroxide ions the it means there are more number of OH^{-} ions into the solution.

Also, it is known that pOH = log \frac{1}{OH^{-}}

Since there is increase in [OH^{-}] concentration so, this means that there will be decrease in the value of pOH.

Also,   pH + pOH = 14

so,                pH = 14 - pOH

                    pH = +ve

Thus, we can conclude that there will be increase in the value of pH.

makvit [3.9K]3 years ago
5 0

Answer:

Its pH value increases.

Explanation:

pH is the measure of alkalinity or acidity of a compound.

pH = - log [H+]

and pH + pOH = 14

where pOH is the measure of basicity of a solution, given by -log[OH-]

As a solution gets more basic that is higher [OH-], the pH increases, and on the other hand, as the pH of a solution decreases by one pH unit, the concentration of H+ increases by ten times.

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Explanation:

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The evidence of the present climate change can be gauged in-

  1. The rise in global average temperature-Earth’s mean temperature has risen by around 1° C ever since the 19th century. 2017 proved to be the warmest year in the record.
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If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

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