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2 years ago

1 mole of phosphoric acid (H3PO4) and 1.75 moles of NaOH are added to the one liter water. Calculate the pH.

1 answer:

6 0

Moles of H⁺ released by each mole of acid = 3
Moles of H⁺ released = 3

Moles of OH⁻ released = 1.75

Moles of H⁺ remaining = 3 - 1.75 = 1.25 mol/dm³

pH = -log[H⁺]
pH = -log(1.25)
pH = -0.1

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$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

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$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

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