The pH of `7 xx 10^(-8) M CH_3COOH" is "(K_w=1 xx 10^(-14))`

Options

A. 8.1
B. 7.9
C. 7.1
D. 6.85

Answer

To find the pH of a 7 x 10^(-8) M solution of acetic acid (CH₃COOH), we will follow these steps:

### Step 1: Understand the Ionization of Acetic Acid
Acetic acid (CH₃COOH) is a weak acid that partially ionizes in water:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set Up the Equilibrium Expression
Let the initial concentration of acetic acid be \( [\text{CH}_3\text{COOH}] = 7 \times 10^{-8} \, \text{M} \). At equilibrium, let \( x \) be the concentration of \( \text{H}^+ \) ions produced. Therefore, at equilibrium:
- \( [\text{H}^+] = x \)
- \( [\text{CH}_3\text{COOH}] = 7 \times 10^{-8} - x \)
- \( [\text{CH}_3\text{COO}^-] = x \)

### Step 3: Consider the Contribution of Water
In a dilute solution, the autoionization of water must also be considered:
\[ \text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^- \]
The concentration of \( \text{H}^+ \) from water is \( 1 \times 10^{-7} \, \text{M} \).

### Step 4: Write the Expression for \( K_w \)
The ion product of water (\( K_w \)) at 25°C is given by:
\[ K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \]
At equilibrium, the total concentration of \( \text{H}^+ \) is:
\[ [\text{H}^+] = x + 1 \times 10^{-7} \]

### Step 5: Set Up the Equation
Substituting into the \( K_w \) expression:
\[ K_w = (x + 1 \times 10^{-7})x \]
Setting this equal to \( 1 \times 10^{-14} \):
\[ (x + 1 \times 10^{-7})x = 1 \times 10^{-14} \]

### Step 6: Solve the Quadratic Equation
This expands to:
\[ x^2 + 1 \times 10^{-7}x - 1 \times 10^{-14} = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1 \times 10^{-7}, c = -1 \times 10^{-14} \):
\[ x = \frac{-1 \times 10^{-7} \pm \sqrt{(1 \times 10^{-7})^2 - 4(1)(-1 \times 10^{-14})}}{2(1)} \]

Calculating the discriminant:
\[ (1 \times 10^{-7})^2 + 4 \times 1 \times 10^{-14} = 1 \times 10^{-14} + 4 \times 10^{-14} = 5 \times 10^{-14} \]

Now substituting back:
\[ x = \frac{-1 \times 10^{-7} \pm \sqrt{5 \times 10^{-14}}}{2} \]
Calculating \( \sqrt{5 \times 10^{-14}} \approx 7.07 \times 10^{-7} \):
\[ x = \frac{-1 \times 10^{-7} + 7.07 \times 10^{-7}}{2} \approx 3.035 \times 10^{-7} \, \text{M} \]

### Step 7: Calculate Total \( [\text{H}^+] \)
Now we can find the total \( [\text{H}^+] \):
\[ [\text{H}^+] = x + 1 \times 10^{-7} \approx 3.035 \times 10^{-7} + 1 \times 10^{-7} = 4.035 \times 10^{-7} \, \text{M} \]

### Step 8: Calculate pH
Finally, calculate the pH:
\[ \text{pH} = -\log([H^+]) = -\log(4.035 \times 10^{-7}) \approx 6.39 \]

### Final Answer
The pH of the 7 x 10^(-8) M CH₃COOH solution is approximately **6.39**.