Calculate the pH of a 0.60 mol L -1 H 2 SO 3 solution that has the stepwise diss

Question

Calculate the pH of a 0.60 mol L-1 H2SO3 solution that has the stepwise dissociation constants Ka1 = 1.5 × 10-2 and Ka2 = 6.3 × 10-8.

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At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

2ICl(g) I2(g) + Cl2(g)

What is the equilibrium concentration of ICl if 0.45 mol of I2 and 0.45 mol of Cl2 are initially mixed in a 2.0 L flask?

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

2ICl(g) I2(g) + Cl2(g)

What is the equilibrium concentration of ICl if 0.45 mol of I2 and 0.45 mol of Cl2 are initially mixed in a 2.0 L flask?

Explanation / Answer

Since second ionisation constant is too low, assume that all the H+ ions coming from only first ionisation.

[H+] = (Kax C)1/2

= (1.5 x 10 -2 x 0.6 )1/2

= 0.09487

pH = - Log (0.09487) = 1.02

Kc = [I2][Cl2] / [ICl]2

0.11 = (0.225 - x) (0.225 - x) / 2x

0.22 x = 0.0506 + x2 - 0.45x

x2 - 0.67 x + 0.0506 = 0

Applying quadratic equation, x = [- 0.67 +/- (0.4489 - 0.2024)] / 2

= 0.08675

Therefore, equilibrium concentration of ICl = 2 x 0.08675 = 0.1735 = 0.17 M

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