Roger Smith

2022-01-09

A 0.185 M solution of a weak acid (HA) has a pH of 2.95. Calculate the acid ionization constant (K3) for the acid.

godsrvnt0706

Step 1
Use the definition of pH to compute the equilibrium hydronium concentration of this solution.
$pH=-\mathrm{log}\left[{H}_{3}{O}^{+}\right]$
$⇒\left[{H}_{3}{O}^{+}\right]={10}^{-pH}$
$={10}^{-2.95}$
$=1.1220×{10}^{-3}M$
Step 2
Create an ICE Table based on the given values, where I ''stands for the initial concentrations, C' stands for the changes in concentration, and'' E' stands for the equilibrium concentrations. Use multiples of z based on the stoichiometric coefficients in the balanced equilibrium equation to represent the changes in concentration in terms of x. Reactants decrease, so their changes are given negative values. Products increase, so their changes are given positive values. Keep in mind that the concentrations of liquids are never included in an ICE table. Note that the initial concentration of hydronium is $1.0×{10}^{-7}$ which is approximated to 0M.
$\begin{array}{|cccccccc|}\hline & HA\left(aq\right)& +& {H}_{2}O\left(I\right)& ⇌& {A}^{-}\left(aq\right)& +& {H}_{2}{O}^{+}\left(aq\right)\\ I& 0.185& & -& & 0& & \approx 0\\ C& -x& & -& & +x& & +z\\ E& 0.185-x& & -& & x& & 1.220×{10}^{-3}\\ \hline\end{array}$
Step 3
Sum the column under ${H}_{3}{O}^{+}$ to solve for x.
$0+x=1.1220×{10}^{-3}$
$x=1.1220×{10}^{-3}$
Step 4
Determine the equilibrium concentrations for all substances.
$\left[HA\right]=0.185-x$
$=0.185-1.1220×{10}^{-3}$
$=0.18388M$
$\left[{A}^{-}\right]=x$
$=1.1220×{10}^{-3}M$
$\left[{H}_{3}{O}^{+}\right]=1.1220×{10}^{-3}M$
Step 5
Calculate the equilibrium constant from the equilibrium constant expression.
${K}_{a}=\frac{\left[{A}^{-}\right]\left[{H}_{3}{O}^{+}\right]}{\left[HA\right]}$
$=\frac{\left(1.1220×{10}^{-3}\right)\left(1.1220×{10}^{-3}\right)}{0.18388}$
$=6.8×{10}^{-6}$

Gerald Lopez

Step 1
Given:
The amount of acid present $HA$ is $0.185M.$
The $pH$ of the given acid is 2.95
Step 2
The formula for the answer is,
$pH=-\mathrm{log}\left[{H}_{3}{O}^{+}\right]$
Change the pH value in the formula above.
$2.95=-\mathrm{log}\left[{H}_{3}{O}^{+}\right]$
${10}^{-2.95}=\left[{H}_{3}{O}^{+}\right]$
$\left[{H}_{3}{O}^{+}\right]=1.122×{10}^{-3}M$
The following is the ICE table for the acid's dissociation reaction:

Step 3
The expression for ${K}_{a}$ is shown below.
${K}_{a}=\frac{\left[{H}_{3}{O}^{+}\right]×\left[{A}^{-}\right]}{\left[HA\right]}$
It is seen from the ICE table that $\left[{H}_{3}{O}^{+}\right]=\left[{A}^{-}\right]=x=1.122×{10}^{-3}$
$\left[HA\right]=0.185-1.122×{10}^{-3}=0.18388$
Replace the values in the equation above.
${K}_{a}=\frac{x×x}{0.185-x}$
$=\frac{1.122×{10}^{-3}×1.122×{10}^{-3}}{0.18388}$
$=6.8×{10}^{-6}$
The ${K}_{a}$ of given acid is $6.8×{10}^{-6}$

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