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# Answer to Question #190351 in Chemistry for Kenan

Question #190351

Calculate the concentration of H, OH and pH and pOH of an aqueous solution of calcium hydroxide with a concentration of 0.005 mol / dm3 assuming complete dissociation

1
2021-05-09T00:00:32-0400

Solution:

calcium hydroxid - Ca(OH)2

Dissociation equation for Ca(OH)2:

Ca(OH)2 → Ca2+ + 2OH-

According to the equation above: CM[Ca(OH)2] = [OH-]/2

[OH-] = 2 × CM[Ca(OH)2] = 2 × 0.005 M = 0.01 M

[OH-] = 0.01 M

We can convert between [OH-] and pOH using the following equation:

pOH = - log[OH-]

pOH = - log(0.01) = 2

pOH = 2

For any aqueous solution at 25âˆ˜C:

pH + pOH = 14

pH = 14 - pOH = 14 - 2 = 12

pH = 12

We can convert between pH and [H+] using the following equation:

pH = - log[H+]

[H+] = 10-pH = 1×10-12 M

[H+] = 1×10-12 M

[H+] = 1×10-12 M;

[OH-] = 0.01 M;

pH = 12;

pOH = 2.

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