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Yuki888 [10]
3 years ago
10

When [OH-] changes from 10-9 to 10-11 in a solution: Kw = [H+][OH-] What happens to [H+]?

Chemistry
2 answers:
victus00 [196]3 years ago
8 0
Kw is a constant number under same temperature. So when [OH-] decrease from 10^-9 to 10^-1, the [H+] will increase. So the answer is a. increases.
uranmaximum [27]3 years ago
4 0

Answer:

[H⁺] increases

Explanation:

Kw = [H⁺][OH⁻]

[OH⁻] = 10⁻⁹

Kw = 1 x 10⁻¹⁴

[H⁺] = Kw / [OH⁻] = 1 x 10⁻¹⁴ / 10⁻⁹= 10⁻⁵

[OH⁻] = 10⁻¹¹

Kw = 1 x 10⁻¹⁴

[H⁺] = Kw / [OH⁻] = 1 x 10⁻¹⁴ / 10⁻¹¹= 10⁻³

As the concentration of OH⁻decreases, the concentration of H⁺ increases.

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Answer:

Option D. 30 mL.

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above,

The mole ratio of the acid, nA = 1

The mole ratio of the base, nB = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base, KOH (Vb) =.?

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Molarity of acid, HNO3 (Ma) = 1.5M

Step 3:

Determination of the volume of the base, KOH needed for the reaction. This can be obtained as follow:

MaVa / MbVb = nA/nB

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Cross multiply

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Divide both side by 0.5

Vb = (1.5 x 10) /0.5

Vb = 30mL

Therefore, the volume of the base, KOH needed for the reaction is 30mL.

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