Question

The pH of an aqueous solution of 8.80x102 M hydroselenic acid, H, Se (aq), is For H,Se K 1 = 1.30-10-4 and K 2 = 1.00*10-11

Answer 1

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Given that, Initial concentration of H₂Se is

8.8 x 10-2 M = 0.088 M

From ICE table:

Hse + H+ H Se = Initial: 0.088 Change: -x Final: 0.088 - x +x x

We know that,

Kai = [H se-]-[H+] (H2Se 2. (0.088 - )

Given that,

Kai = 1.3 x 10-4

"(0.088 - 0) = 1.3 x 10-4

.

:..? = 1.3 x 10-4 (0.088 - 1)

Solving for x from quadratic equation, we get,

I = [H+1 = 0.0033179314 M

Therefore, pH is given by,

pH = -log[H+1= -log (0.0033179314 M)

:.pH = -2.479)

..pH = 2.48

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