## Chemistry: Molecular Approach (4th Edition)

- Calculate the molar mass: $C_8H_{10}N_4O_2$ : ( 1.008 $\times$ 10 )+ ( 12.01 $\times$ 8 )+ ( 14.01 $\times$ 4 )+ ( 16.00 $\times$ 2 )= 194.2 g/mol - Use the information as conversion factors to find the molarity of this solution: $$\frac{ 455 mg \space C_8H_{10}N_4O_2 }{1 \space L \space solution} \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mol \space C_8H_{10}N_4O_2 }{ 194.2 \space g \space C_8H_{10}N_4O_2 } = 2.34 \times 10^{-3} \space M$$ - Calculate the $K_b$ $$K_b = 10^{-pK_b} = 3.98 \times 10^{-11}$$ 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ C_8H_{10}N_4O_2 ]& [ C_8H_{11}N_4O{_2}^{+} ]& [ OH^- ]\\ Initial& 2.34 \times 10^{-3} & 0 & 0 \\ Change& -x& +x& +x\\ Equilibrium& 2.34 \times 10^{-3} -x& 0 +x& 0 +x\\ \end{vmatrix}$$ 2. Write the expression for $K_b$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_b = \frac{[Products]}{[Reactants]} = \frac{[ C_8H_{11}N_4O{_2}^{+} ][ H^+ ]}{[ C_8H_{10}N_4O_2 ]}$$ $$K_b = \frac{(x)(x)}{[ C_8H_{10}N_4O_2 ]_{initial} - x}$$ 3. Assuming $2.34 x 10^{-3} \gt\gt x$: $$K_b = \frac{x^2}{[ C_8H_{10}N_4O_2 ]_{initial}}$$ $$x = \sqrt{K_b \times [ C_8H_{10}N_4O_2 ]_{initial}} = \sqrt{ 3.98 \times 10^{-11} \times 2.34 \times 10^{-3} }$$ $x = 3.05 \times 10^{-7}$ 4. Test if the assumption was correct: $$\frac{ 3.05 \times 10^{-7} }{ 2.34 \times 10^{-3} } \times 100\% = 0.013 \%$$ 5. Thus, it is correct to say that $x = 3.05 \times 10^{-7}$ 6. $[OH^-] = x = 3.05 \times 10^{-7}$ 7. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.05 \times 10^{-7} } = 3.3 \times 10^{-8} \space M$$ $$pH = -log[H_3O^+] = -log( 3.3 \times 10^{-8} ) = 7.48$$