Question

in an aqueous solution of monoprotic acid (ka=10-5)is diluted 10 times then Ka

Answer 1

Answer:

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Explanation:

Calculate the [H+] using the Ka equation

If the wreak acid is HA. This will produce a low concentration of H+ and A- ions . There will be undissociated HA in solution

Ka = [H+] [A-] / [HA]

Because [H+] = [A-] we can write [H+] [A-] as [H+]² and because dissociation is small we can use 0.0933 for [HA]

Substitute

1.0*10^-5 = [H+]² / 0.0933

[H+]² = 0.0933 ( 1*10^-5)

[H+]² = 9.33*10–7

[H+] = 9.66*10^-4 M

pH = - log [H+]

pH = - log ( 9.66*10^-4)

pH = 3.015 ( answer to 3 significant digits. )