Answer
The pH of this solution is equal to 7.48
Work Step by Step
- 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:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.05 \times 10^{-7} } = 3.3 \times 10^{-8} \space M$$
$$pH = -log[H_3O^+] = -log( 3.3 \times 10^{-8} ) = 7.48 $$
