Answer
$pH = 7.48$
Work Step by Step
1. Calculate the molar mass:
12.01* 8 + 1.01* 10 + 14.01* 4 + 16* 2 = 194.22g/mol
2. Calculate the number of moles
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{0.455}{ 194.22}$
$n(moles) = 2.34\times 10^{- 3}$
3. Find the concentration in mol/L:
$2.34 \times 10^{-3}$ mol in 1L: $2.34 \times 10^{-3} M$
- Calculate the $K_b$ using the $pK_b$
$K_b = 10^{-pKb}$
$K_b = 10^{- 10.4}$
$K_b = 3.98 \times 10^{- 11}$
- We have these concentrations at equilibrium:
-$[OH^-] = [C_8H_{11}N_4{O_2}^+] = x$
-$[C_8H_{10}N_4O_2] = [C_8H_{10}N_4O_2]_{initial} - x = 2.34 \times 10^{- 3} - x$
For approximation, we consider: $[C_8H_{10}N_4O_2] = 2.34 \times 10^{- 3}M$
- Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][C_8H_{11}N_4{O_2}^+]}{ [C_8H_{10}N_4O_2]}$
$Kb = 3.98 \times 10^{- 11}= \frac{x * x}{ 2.34\times 10^{- 3}}$
$Kb = 3.98 \times 10^{- 11}= \frac{x^2}{ 2.34\times 10^{- 3}}$
$ 9.33 \times 10^{- 14} = x^2$
$x = 3.05 \times 10^{- 7}$
Percent ionization: $\frac{ 3.05 \times 10^{- 7}}{ 2.34\times 10^{- 3}} \times 100\% = 0.013\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [C_8H_{11}N_4{O_2}^+] = x = 3.05 \times 10^{- 7}M $
$[C_8H_{10}N_4O_2] \approx 2.34 \times 10^{-3}M$
$pOH = -log[OH^-]$
$pOH = -log( 3.05 \times 10^{- 7})$
$pOH = 6.52$
$pH + pOH = 14$
$pH + 6.52 = 14$
$pH = 7.48$
