Answer
Hydronium ion concentration:
$[H_3O^+] = 1.1 \times 10^{- 8}M$
$pH = 7.96$
Work Step by Step
Analyze each ion that forms that salt:
$Na^+$: Conjugate acid of $NaOH$ (strong base). Therefore, it will not affect the pH of the water.
$HCO_2^-$: Conjugate base of a weak acid $(HCO_2H)$.
Calculate the pH for a $0.015M$ $HCO_2^-$ solution.
- Since $HCO_2^-$ is the conjugate base of $HCO_2H$ , we can calculate its $K_b$ using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 4} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 4}}$
$K_b = 5.6\times 10^{- 11}$
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HCO_2^-(aq) + H_2O(l) \lt -- \gt HCO_2H(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HCO_2H$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HCO_2H] = 0 + x = x$
-$[HCO_2^-] = [HCO_2^-]_{initial} - x $
For approximation, we are going to consider $[HCO_2^-]_{initial} = [HCO_2^-]$
2. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HCO_2H]}{ [HCO_2^-]}$
$Kb = 5.6 \times 10^{- 11}= \frac{x * x}{ 0.015}$
$Kb = 5.6 \times 10^{- 11}= \frac{x^2}{ 0.015}$
$x^2 = 5.6 \times 10^{-11} \times 0.015$
$x = \sqrt { 5.6 \times 10^{-11} \times 0.015} = 9.2 \times 10^{-7}$
Percent ionization: $\frac{ 9.2 \times 10^{- 7}}{ 0.015} \times 100\% = 0.0061\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HCO_2H] = x = 9.2 \times 10^{- 7}M $
$[HCO_2^-] \approx 0.015M$
3. Calculate the hydronium ion concentration:
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 9.2 \times 10^{- 7} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 9.2 \times 10^{- 7}}$
$[H_3O^+] = 1.1 \times 10^{- 8}M$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 1.1 \times 10^{- 8})$
$pH = 7.96$
