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motikmotik
2 years ago
13

How will the concentration of H+ and OH− ions change when a substance with a pH 11.4 is added to water?

Chemistry
2 answers:
PolarNik [594]2 years ago
4 0

Answer is: H+ will increase and OH− will decrease.

pH = 11.4.

pH = -log[H+].

[H+] = 10∧(-pH).

[H+] = 10∧(-11.4).

[H+] = 4·10⁻¹² mol/L.

[H+] · [OH-] = 1·10⁻¹⁴ mol²/dm⁶.

[OH-] = 0.0025 mol/L.

pH value (potential of hydrogen - [H+]) is a logarithmic scale that specify the acidity or basicity of an aqueous solution.

When pH is greater than seven, aqueous solution is basic, below seven is acidic and when pH is equal seven, solution is neutral.



Gekata [30.6K]2 years ago
4 0

Answer : The correct option is, H^+ will decrease and OH^- will increase

Explanation :

As we know that the water is made up of the H^+ and OH^- ions. When a substance with a pH 11.4 that means the solution is basic in nature is added to water, the concentration of OH^- ions increase and the concentration of H^+ ions decreases.

In the basic solution, the concentration of OH^- ions are more than the concentration of H^+ ions. So, when a base is added to the water the more OH^- ions dissociates in the water as compared to the H^+ ions.

Hence, the correct option is, H^+ will decrease and OH^- will increase

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3 years ago
Write a molecular equation for the precipitation reaction that occurs (if any) when the following solutions are mixed. If no rea
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Answer :

(A) The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

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Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

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Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

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Explanation :

Molecular equation : It is defined as a balanced chemical equation where the ionic compounds are expressed in the form of molecules rather than component of ions.

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The insoluble salt that settle down in the solution is known an precipitate.

Part A  : potassium carbonate and lead(II) nitrate

The balanced molecular equation will be:

K_2CO_3(aq)+Pb(NO_3)_2(aq)\rightarrow 2KNO_3(aq)+PbCO_3(s)

In this reaction, lead carbonate is an insoluble salt and potassium nitrate is a soluble solution.

Part B : lithium sulfate and lead(II) acetate

The balanced molecular equation will be:

Li_2SO_4(aq)+Pb(CH_3COOH)_2(aq)\rightarrow 2LiCH_3COOH(aq)+PbSO_4(s)

In this reaction, lead sulfate is an insoluble salt and lithium acetate is a soluble solution.

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The balanced molecular equation will be:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

In this reaction, Cuprous sulfide is an insoluble salt and sodium nitrate is a soluble solution.

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The balanced molecular equation will be:

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow 2KNO_3(aq)+SrI_2(aq)

In this reaction, strontium iodide and potassium nitrate are soluble solution.

Sr(NO_3)_2(aq)+2KI(aq)\rightarrow \text{No reaction}

6 0
3 years ago
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Potash is usually in the form of potassium salts such potassium chloride and  potassium sulphate.  The potash  is mined then taken to the factory where it is crushed and purified  by removing such impurities as clay.

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5 0
3 years ago
An unknown triprotic acid (H3A) is titrated with NaOH. After the titration Ka1 is determined to be 0.0013 and Ka2 is determined
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From the second Ka, you can calculate pKa = -log (Ka2) = 6.187

The pH at the second equivalence point (8.181) will be the average of pKa2  and pKa3. So,

8.181 = (6.187 + pKa3) / 2

Solving gives pKa3 = 10.175, and Ka3 = 10^-pKa3 = 6.68 X 10^-11

7 0
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