Question

Calculate the concentration of hydrogen ion in 0.62M NaOH solution

Answer 1

Answer:

K

w = [H

3

O

+][OH-] = 1.0 x 10-14

[H

3

O

+] x 0.62 M = 1.0 x 10-14

[H

+] = (1.0 x 10-14/ 0.62) = 1.6 x 10-14

M

Explanation:

hope it helps you

Answer 2

Answer:

The concentration of hydrogen ions in a 0.62M NaOH solution is 1.61 × $10^{-14}$ M.

Explanation:

Given that,

The molarity of the NaOH solution is 0.62M

So,

0.62 M NaOH ==> 0.62 M Na⁺ + 0.62 M OH⁻

In a 0.62 M NaOH solution, the concentration of OH ions is equal to the NaOH concentration, which is 0.62 M.

[OH⁻] = 0.62M

From the equation for the self-ionization of water, we get

H₂O ⇌ H⁺  + OH⁻

The hydrogen ions (H⁺) concentration in the NaOH solution can be calculated using the hydroxide ions (OH⁻) concentration in the NaOH solution and the equilibrium constant (Kw).

Kw = 1.0 × $10^{-14}$

Kw = [H⁺][OH⁻]

[H⁺] = Kw / [OH⁻]

      = 1.0 × $10^{-14}$ / 0.62

      = 1.61 × $10^{-14}$ M

The concentration of H⁺ ions = 1.61 × $10^{-14}$ M.

Therefore, the concentration of H⁺ ions in a 0.62M NaOH solution is 1.61 × $10^{-14}$ M.

To know more about the concentration of hydrogen ions, visit the link below:

https://brainly.in/question/2946067

To know more about the molarity, visit the link below:

https://brainly.in/question/1159016

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