Chapter 15 - Section 15.3 - Equilibrium Expressions - Checkpoint - Page 678: 15.3.3

c.$$K_c = 1.2 \times 10^{-1}$$

Work Step by Step

1. Multiply all coefficients of the first reaction by 2, so find the value of Kc to the power of 2. $$2HF(aq) \leftrightharpoons 2H^+(aq) + 2F^-(aq)$$ $$K_c = (6.8 \times 10^{-4})^2 = 4.6 \times 10^{-7}$$ 2. Reverse the second reaction, so find the inverse of Kc. $$2H^+(aq) + C_2{O_4}^{2-}(aq) \leftrightharpoons H_2C_2O_4(aq)$$ $$K_c = \frac{1}{3.8 \times 10^{-6}} = 2.6 \times 10^5$$ 3. Thus: $$2 HF(aq) + C_2O_4^{2-} \leftrightharpoons 2F^-(aq) + H_2C_2O_4(aq)$$ $$K_c = 4.6 \times 10^{-7} \times 2.6 \times 10^5 = 1.2 \times 10^{-1}$$

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