Calculate: (H+) of 0.02 M acid solution whose pH is 4.6.

Question

Question

Answer 1

Answer #1

pH = -log [H₃O]⁺ = -log [H]⁺

by putting the value

4.6 = -log [H⁺]

[H⁺]= 10 ^-4.6

= 0.000025 M

For 0.02 M acid [H⁺] equal to

= 0.02 x 0.000025

= 5 x 10 ^-7 M

Therefore (H+) of 0.02 M acid solution whose pH is 4.6 = 5 x 10 ^-7 M

= 2.5 x 10^-5 M

Answer 2

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