What is the concentration of S2- in a 0.25 M solution of NaHS? Ka of HS1-...

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Question

What is the concentration of S^2- in a 0.25 M solution of NaHS? K_a of HS^1- = 1.0 x 10^-19

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Answer 1

Answer #1

HS-(aq) <---> S2-(aq) + H+ (aq) ,

at equilibrium [HS-] = 0.25-X , [S2-] = [H+] = X

Ka1 = [S2-] [H+] /[HS-]

10^-19 = ( X)(X) / ( 0.25-X)

10^-19 = X^2/( 0.25-X) ( 0.25-X is nearly equal to 0.25 since we get X very less value)

X^2 = 0.25 x 10^-19

X = [H+] = 1.58 x 10^-10 M

pH = -log ( 1.58x10^-10) = 9.8

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Answer 2

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