Answer
$[OH^-] = [HN_3] = 2.3 \times 10^{- 6}M $
$[N_3^-] = 0.010M$
$ [Na^+] = 0.010M$
$[H_3O^+] = 4.3 \times 10^{- 9}M$
Work Step by Step
1. Analyze each ion that forms this salt:
$Na^+$: Conjugate pair of a strong base (NaOH). So, it will not affect the pH of the solution.
$N_3^-$: Conjugate pair of a weak acid. So, it will affect the pH of the solution, because it acts as a weak base.
- Find the $[OH^-]$ for a $0.010M $ $N_3^-$ solution:
2. Since $N_3^-$ is the conjugate base of $HN_3$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.9\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.9\times 10^{- 5}}$
$K_b = 5.3\times 10^{- 10}$
3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$N_3^-(aq) + H_2O(l) \lt -- \gt HN_3(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HN_3$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HN_3] = 0 + x = x$
-$[N_3^-] = [N_3^-]_{initial} - x $
For approximation, we are going to consider $[N_3^-]_{initial} = [N_3^-]$
4. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HN_3]}{ [N_3^-]}$
$Kb = 5.3 \times 10^{- 10}= \frac{x * x}{ 0.010}$
$Kb = 5.3 \times 10^{- 10}= \frac{x^2}{ 0.010}$
$x^2 = 5.3 \times 10^{-10} \times 0.010$
$x = \sqrt { 5.3 \times 10^{-10} \times 0.010} = 2.3 \times 10^{-6}$
Percent ionization: $\frac{ 2.3 \times 10^{- 6}}{ 0.01} \times 100\% = 0.023\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HN_3] = x = 2.3 \times 10^{- 6}M $
$[N_3^-] \approx 0.010M$
5. Calculate the $[H_3O^+]$:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.3 \times 10^{- 6} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.3 \times 10^{- 6}}$
$[H_3O^+] = 4.3 \times 10^{- 9}M$
6. Since $Na^+$ didn't react with water, its concentration didn't change from the initial one:
$0.010M (NaN_3) = 0.010M(Na^+)$
