Answer
a. pH = 2.00;
The major species are: $HNO_2$ and $H_2O$.
a. pH = 2.68;
The major species are: $CH_3CO_2H$ and $H_2O$.
Work Step by Step
a.
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$HNO_2(aq) + H_2O(l) \lt -- \gt NO_2^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $NO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [NO_2^-] = 0 + x = x$
-$[HNO_2] = [HNO_2]_{initial} - x$
For approximation, we are going to consider $[HNO_2]_{initial} = [HNO_2]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][NO_2^-]}{ [HNO_2]}$
$Ka = 4 \times 10^{- 4}= \frac{x * x}{ 0.250}$
$Ka = 4 \times 10^{- 4}= \frac{x^2}{ 0.250}$
$x^2 = 0.250 \times 4 \times 10^{-4} $
$x = \sqrt { 0.250 \times 4 \times 10^{-4}} = 0.01 $
Percent dissociation: $\frac{ 1 \times 10^{- 2}}{ 0.25} \times 100\% = 4\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [NO_2^-] = x = 1 \times 10^{- 2}M $
And, since 'x' has a very small value (compared to the initial concentration): $[HNO_2] \approx 0.250M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.010)$
$pH = 2.00$
The species with a significant concentration in this solution are: $HNO_2$ and $H_2O$.
-----------------------
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:
For the reaction:
$CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3CO_2^-(aq) + H_3O^+(aq) $
Remember:
Reactants get a '-x', and products get a '+x'
The initial concentration of $CH_3CO_2^-$ is : 0 M, and the same for $[H_3O^+]$ is $\approx$ 0 M
-$[H_3O^+] = [CH_3CO_2^-] = 0 + x = x$
-$[CH_3CO_2H] = [CH_3CO_2H]_{initial} - x$
For approximation, we are going to consider $[CH_3CO_2H]_{initial} = [CH_3CO_2H]$
2. Now, use the Ka value and equation to find the 'x' value.
$Ka = \frac{[H_3O^+][CH_3CO_2^-]}{ [CH_3CO_2H]}$
$Ka = 1.8 \times 10^{- 5}= \frac{x * x}{ 0.250}$
$Ka = 1.8 \times 10^{- 5}= \frac{x^2}{ 0.250}$
$x^2 = 0.250 \times 1.8 \times 10^{-5} $
$x = \sqrt { 0.250 \times 1.8 \times 10^{-5}} = 2.1 \times 10^{-3} $
Percent dissociation: $\frac{ 2.1 \times 10^{- 3}}{ 0.250} \times 100\% = 0.85\%$
%dissociation < 5% : Right approximation.
Therefore: $[H_3O^+] = [CH_3CO_2^-] = x = 2.1 \times 10^{- 3}M $
And, since 'x' has a very small value (compared to the initial concentration): $[CH_3CO_2H] \approx 0.250M$
3. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 2.1 \times 10^{- 3})$
$pH = 2.68$
The species with a significant concentration in this solution are: $CH_3CO_2H$ and $H_2O$.