Answer
a. pH = 0.939
b. pH = 1.07
c. pH = 2.19
d. pH = 3.02
Work Step by Step
a.
- HBr is a strong acid that will completely ionize, producing 0.115 M of $H_3O^+$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.125 & 0 & 0.115 \\
Change& -x& +x& +x\\
Equilibrium& 0.125 -x& 0 +x& 0.115 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$
$$K_a = \frac{(x)( 0.115 + x)}{[ HCHO_2 ]_{initial} - x}$$
3. Assuming $ 0.125 \space and \space 0.115 \gt\gt x:$
$$K_a = \frac{(x)( 0.115 )}{[ HCHO_2 ]_{initial}}$$
$$x = \frac{K_a \times [ HCHO_2 ]_{initial}}{ 0.115 } = \frac{ 1.8 \times 10^{-4} \times 0.125 }{ 0.115 } $$
$x = 2.0 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 2.0 \times 10^{-4} }{ 0.125 } \times 100\% = 0.16 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 2.0 \times 10^{-4} $
6. $$[H_3O^+] = x + [H_3O^+]_{initial} = 2.0 \times 10^{-4} + 0.115 = 0.115 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log(0.115) = 0.939 $$
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b.
- $HNO_3$ is a strong acid that will completely ionize, producing 0.085 M of $H_3O^+$
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HNO_2 ]& [ N{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.150 & 0 & 0.085 \\
Change& -x& +x& +x\\
Equilibrium& 0.150 -x& 0 +x& 0.085 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ N{O_2}^- ][ H_3O^+ ]}{[ HNO_2 ]}$$
$$K_a = \frac{(x)( 0.085 + x)}{[ HNO_2 ]_{initial} - x}$$
3. Assuming $ 0.150 \space and \space 0.085 \gt\gt x:$
$$K_a = \frac{(x)( 0.085 )}{[ HNO_2 ]_{initial}}$$
$$x = \frac{K_a \times [ HNO_2 ]_{initial}}{ 0.085 } = \frac{ 4.6 \times 10^{-4} \times 0.150 }{ 0.085 } $$
$x = 8.1 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 8.1 \times 10^{-4} }{ 0.150 } \times 100\% = 0.54 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 8.1 \times 10^{-4} $
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 8.1 \times 10^{-4} + 0.085 = 0.086 $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 0.086 ) = 1.07 $$
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c.
- Formic acid $(HCHO_2)$ is the stronger of these two; start by calculating the hydronium concentration of a pure formic acid with this molarity:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HCHO_2 ]& [ CH{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.185 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.185 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ CH{O_2}^- ][ H_3O^+ ]}{[ HCHO_2 ]}$$
$$K_a = \frac{(x)(x)}{[ HCHO_2 ]_{initial} - x}$$
3. Assuming $ 0.185 \gt\gt x:$
$$K_a = \frac{x^2}{[ HCHO_2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HCHO_2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-4} \times 0.185 }$$
$x = 5.8 \times 10^{-3} $
4. Test if the assumption was correct:
$$\frac{ 5.8 \times 10^{-3} }{ 0.185 } \times 100\% = 3.1 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 5.8 x 10^{-3} $
6. $$[H_3O^+] = x = 5.8 \times 10^{-3} $$
- Now, calculate the hydronium concentration of the mixture when we add the weaker acid:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_2H_3O_2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.225 & 0 & 5.8 x 10^{-3} \\
Change& -x& +x& +x\\
Equilibrium& 0.225 -x& 0 +x& 5.8 x 10^{-3} +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H_3O^+ ]}{[ HC_2H_3O_2 ]}$$
$$K_a = \frac{(x)( 5.8 \times 10^{-3} + x)}{[ HC_2H_3O_2 ]_{initial} - x}$$
3. Assuming $ 0.225 \space and \space 5.8 x 10^{-3} \gt\gt x:$
$$K_a = \frac{(x)( 5.8 \times 10^{-3} )}{[ HC_2H_3O_2 ]_{initial}}$$
$$x = \frac{K_a \times [ HC_2H_3O_2 ]_{initial}}{ 5.8 \times 10^{-3} } = \frac{ 1.8 \times 10^{-5} \times 0.225 }{ 5.8 \times 10^{-3} } $$
$x = 7.0 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 7.0 \times 10^{-4} }{ 0.225 } \times 100\% = 0.31 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 7.0 \times 10^{-4} $
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 7.0 \times 10^{-4} + 5.8 \times 10^{-3} = 6.5 \times 10^{-3} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 6.5 \times 10^{-3} ) = 2.19 $$
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d.
- Since the Ka for hydrocyanic acid is very small compared to the same for acetic acid, we can ignore the presence of hydrocyanic acid in the mixture:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HC_2H_3O2 ]& [ C_2H_3{O_2}^- ]& [ H_3O^+ ]\\
Initial& 0.050 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.050 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C_2H_3{O_2}^- ][ H_3O^+ ]}{[ HC_2H_3O2 ]}$$
$$K_a = \frac{(x)(x)}{[ HC_2H_3O2 ]_{initial} - x}$$
3. Assuming $ 0.050 \gt\gt x:$
$$K_a = \frac{x^2}{[ HC_2H_3O2 ]_{initial}}$$
$$x = \sqrt{K_a \times [ HC_2H_3O2 ]_{initial}} = \sqrt{ 1.8 \times 10^{-5} \times 0.050 }$$
$x = 9.5 \times 10^{-4} $
4. Test if the assumption was correct:
$$\frac{ 9.5 \times 10^{-4} }{ 0.050 } \times 100\% = 1.9 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 9.5 x 10^{-4} $
6. $$[H_3O^+] = x = 9.5 \times 10^{-4} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-4} ) = 3.02 $$
