Answer
a.
$[H_3O^+] = [OH^-] = 1.0 \times 10^{-7}M$
$pH = 7.00$
b.
$[H_3O^+] = 4.2 \times 10^{-10}M$
$[OH^-] = 2.4 \times 10^{-5}M$
$pH = 9.38$
Work Step by Step
a.
1. Analyze each ion that forms this salt:
$K^+$: Conjugate pair of a strong base (KOH). So, it will not affect the pH of the solution.
$Cl^-$: Conjugate pair of a strong acid (HCl). So, it will not affect the pH of the solution.
2. Therefore, the $[OH^-], [H_3O^+]$ and $pH$ will not be affected. So the values for that solution are equal to a pure water solution.
$[H_3O^+] = [OH^-] = 1.0 \times 10^{-7}M$
$pH = 7.00$
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b.
1. Analyze each ion that forms this salt:
$K^+$: Conjugate pair of a strong base (KOH). So, it will not affect the pH of the solution.
$C_2H_3O_2^-$: Conjugate pair of a weak acid. So, it will act as a weak base.
- Calculate the $[OH^-]$ considering a $1.0M$ $C_2H_3O_2^-$ solution:
- Since $C_2H_3O_2^-$ is the conjugate base of $HC_2H_3O_2$ , we can calculate its kb by using this equation:
$K_a * K_b = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_b = 10^{-14}$
$K_b = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_b = 5.6\times 10^{- 10}$
3. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$C_2H_3O_2^-(aq) + H_2O(l) \lt -- \gt HC_2H_3O_2(aq) + OH^-(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $HC_2H_3O_2$ is : 0 M, and the same for $[OH^-]$ is $\approx$ 0 M
-$[OH^-] = [HC_2H_3O_2] = 0 + x = x$
-$[C_2H_3O_2^-] = [C_2H_3O_2^-]_{initial} - x $
For approximation, we are going to consider $[C_2H_3O_2^-]_{initial} = [C_2H_3O_2^-]$
4. Now, use the Kb value and equation to find the 'x' value.
$Kb = \frac{[OH^-][HC_2H_3O_2]}{ [C_2H_3O_2^-]}$
$Kb = 5.6 \times 10^{- 10}= \frac{x * x}{ 1}$
$Kb = 5.6 \times 10^{- 10}= \frac{x^2}{ 1}$
$x^2 = 5.6 \times 10^{-10} \times 1$
$x = \sqrt { 5.6 \times 10^{-10} \times 1} = 2.4 \times 10^{-5}$
Percent ionization: $\frac{ 2.4 \times 10^{- 5}}{ 1} \times 100\% = 0.0024\%$
%ionization < 5% : Right approximation.
Therefore: $[OH^-] = [HC_2H_3O_2] = x = 2.4 \times 10^{- 5}M $
$[C_2H_3O_2^-] \approx 1M$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.4 \times 10^{- 5} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.4 \times 10^{- 5}}$
$[H_3O^+] = 4.2 \times 10^{- 10}$
5. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 4.2 \times 10^{- 10})$
$pH = 9.38$
