Solutions
- Page ID
- 11056
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution. 
Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.
| HC3H5O2 (aq) |  |
H+(aq) | + C3H5O2- (aq) | |
| initial concentration | 0.2 M | 0 | 0 | |
| change | - x | + x | + x | |
| equilibrium concentration | 0.2 - x | x | x |
Write the equilibrium expression (Ka) and express the Ka in terms of the equilibrium concentrations.
| [H+][C3H5O2-] | (x)(x) | |||
| Ka = | _______________ | = | ___________ | = 1.3 x 10-5 |
| [HC3H5O2] | (0.2 - x) | (from table) |
Solve for x; since the initial concentration is greater than 0.1 and the Ka less than 10-4, we can drop the '- x' term.
| x2 | solving for 'x' gives | ||
| _______ | = 1.3 x 10-5 | x = 1.61 x 10-3 | |
| 0.2 |
Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H+] = [C3H5O2-] = x, [HC3H5O2] = 0.2 - x
[H+] = [C3H5O2-] = 1.61 x 10-3 M, [HC3H5O2] = 1.98 x 10-1 M;
pH = - log [H+] = - log(1.61 x 10-3)
pH = 2.79
2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.
Compare the answer to problem 1. Note which has the higher Ka and lower pH.
This is done the same way as problem 1, using a Ka value of 4.2 x 10-7
| H2CO3 (aq) | |
H+(aq) | + HCO3- (aq) | |
| initial concentration | 0.2 M | 0 | 0 | |
| change | - x | + x | + x | |
| equilibrium concentration | 0.2 - x | x | x |
| [H+][HCO3-] | (x)(x) | |||
| Ka = | _______________ | = | ___________ | = 4.2 x 10-7 |
| [H2CO3] | (0.2 - x) | (from table) |
| x2 | solving for 'x' gives | ||
| _______ | = 4.2 x 10-7 | x = 2.90 x 10-4 | |
| 0.2 |
[H+] = [HCO3-] = 2.90 x 10-4 M, [H2CO3] = 2.0 x 10-1 M; pH = 3.54
The higher the Ka, the lower the pH for the same concentration of weak acid. The higher Ka value tells us we have more dissociation of the weak acid, giving a greater [H+].
3. Calculate the Ka for a 0.3 M solution of HA (weak acid) if the pH = 3.65
First calculate the [H+] concentration from the pH
[H+] = 10-pH; [H+] = 10-3.65; [H+] = 2.24 x 10-4
since 1:1 mole ratio
[H+] = [A-]; [HA] = 0.3 - [H+]
Substitute into the Ka expression
| [H+][A-] | (2.24 x 10-4)(2.24 x 10-4) | ||
| Ka = | ________ | = | ________________________ |
| [HA] | (0.3 - 2.24 x 10-4) |
Ka= 1.67 x 10-7
4. Calculate the pH for a 0.2 M pyridine solution.
| C5H5N (aq) | |
OH-(aq) | + C5H5NH+ (aq) | |
| initial concentration | 0.2 M | 0 | 0 | |
| change | - x | + x | + x | |
| equilibrium concentration | 0.2 - x | x | x |
| [OH-][C5H5NH+] | (x)(x) | |||
| Kb = | ________________ | = | ___________ | = 2.0 x 10-9 |
|
[C5H5N] |
(0.2 - x) | (from table) |
Solve for 'x'
| x2 | solving for 'x' gives | ||
| _______ | = 2.0 x 10-9 | x = 2.0 x 10-5 | |
| 0.2 |
[OH-] = [C5H5NH+] = 2.0 x 10-5 M, [C5H5N] = 2.0 x 10-1 M
Using the following equations: pOH = - log [OH-]; pH + pOH = 14
pOH = 4.70, pH = 9.30
5. Calculate the pH of a 0.34 M HCl solution.
pH = 0.47, strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H+]