## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Self-Assessment Quiz - Page 768: Q13

(b) 11.28

#### Work Step by Step

1. $CN^-$ is the conjugate base of a weak acid; therefore, it is a weak base. 2. Find its $K_b$: $$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{ 4.9 \times 10^{-10} } = 2.0 \times 10^{-5}$$ 3. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ CN^- ]& [ HCN ]& [ OH^- ]\\ Initial& 0.175 & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 0.175 -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 4. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HCN ][ H^+ ]}{[ CN^- ]}$$ $$K_b = \frac{(x)(x)}{[ CN^- ]_{initial} - x}$$ 5. Assuming $0.175 \gt\gt x$: $$K_b = \frac{x^2}{[ CN^- ]_{initial}}$$ $$x = \sqrt{K_b \times [ CN^- ]_{initial}} = \sqrt{ 2.0 \times 10^{-5} \times 0.175 }$$ $x = 1.9 \times 10^{-3}$ 6. Test if the assumption was correct: $$\frac{ 1.9 \times 10^{-3} }{ 0.175 } \times 100\% = 1.1 \%$$ 7. Thus, it is correct to say that $x = 1.9 \times 10^{-3}$ 8. $[OH^-] = x = 1.9 \times 10^{-3}$ 9. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.9 \times 10^{-3} } = 5.3 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 5.3 \times 10^{-12} ) = 11.28$$

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