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What is the ph of a 2. 3 m solution of the weak acid h2s, with a ka of 1. 19×10−7? the equilibrium expression is: h2s(aq) h2o(l)⇋h3o (aq) hs−(aq)

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The equilibrium expression is:

H2S (aq) + H2O (l) ⇋ H3O (aq) + HS (aq)

In order to calculate the pH of equation having aqueous solution of weak acid H2S, let’s assume that molar concentration of the products H3O and HS total released is ‘a’. Now as per given information concentration of weak acid H2S is 2.3 M. Thereby,

Ka = [H3O+] [HS] / [H2S]

1.19 x 10-7 = [a] * [a] / [2.3]

a2 = 1.19 x 10-7 x 2.3

a = 5.24 x 10-4

As per the calculation, the total concentration of the products released (HS) and (H3O) will be 5.24 x 10-4, on that basis, the pH of the weak acid (H2S) can be calculated as follows:

-log (0.000522494) = 3.28

To know more about the pH calculation:

https://brainly.com/question/13105728

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