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3 years ago

Consider the following reaction:

1 answer:

4 0

<span>Kc = [H2S]²*[O2]³ / [H2O]²*[SO2]²
Let x be the moles of H2S formed. Each mole of H2S takes one each mole of H2O and SO2 so after the reactions
[H2O] = 2.8 - x and [SO2] = 2.6 - x also for each mole of H2S, 1.5 moles of O2 are formed, so [O2] = 1.5*x
Kc = x²*(1.5*x)³ / (2.8 - x)²*(2.6 - x)²
Thus use 2.8 - x = 2.8 and 2.6 - x = 2.6 in the above equation for Kc:
Kc = x²*(1.5*x)³ / 2.8²*2.6² = 3.375x^5 / 2.8²*2.6² = 0.06368*x^5
x^5 = 1.3*10^-6 / 0.06368 = 2.0414*10^-5
x = 0.115M </span> hope it helps

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$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

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Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

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Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

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K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

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$pK_{a} = -log K_{a}$

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