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# mar1008 - Table 16.3 page 541 11.1 What is the pH of a 0.5...

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11.1 Table 16.3 page 541

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What is the pH of a 0.5 M HF solution (at 25 0 C)? HF ( aq ) H + ( aq ) + F - ( aq ) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF ( aq ) H + ( aq ) + F - ( aq) Initial ( M ) Change ( M ) Equilibrium ( M ) 0.50 0.00 - x + x 0.50 - x 0.00 + x x x K a = x 2 0.50 - x = 7.1 x 10 -4 K a x 2 0.50 = 7.1 x 10 -4 0.50 – x 0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M 11.2
0.50 – x 0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? K a x 2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation or method of successive approximation. 11.3

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## mar1008 - Table 16.3 page 541 11.1 What is the pH of a 0.5...

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