Determine the pH of 0.265 M HClO solution. The Ka of HClO is 2.9 x 10^-8
answers could be…
a) 0.576
b) 8.11
c) 5.89
d) 4.06
c)3.31
The correct answer and explanation is :
To determine the pH of a 0.265 M HClO (hypochlorous acid) solution, we need to account for the dissociation of HClO in water. HClO is a weak acid, meaning it does not completely dissociate in water. Instead, it establishes an equilibrium between its undissociated form (HClO) and its dissociated ions (H+ and ClO−).
The dissociation reaction of HClO is:
[
\text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^-
]
The equilibrium constant, Ka, is given as:
[
\text{Ka} = 2.9 \times 10^{-8}
]
Step 1: Set up the ICE table (Initial, Change, Equilibrium)
We will use an ICE (Initial, Change, Equilibrium) table to track the concentrations of the species involved in the dissociation.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HClO | 0.265 | -x | 0.265 – x |
| H+ | 0 | +x | x |
| ClO− | 0 | +x | x |
Step 2: Write the expression for Ka
The equilibrium constant expression for this dissociation is:
[
\text{Ka} = \frac{[H^+][ClO^-]}{[HClO]} = \frac{x \cdot x}{0.265 – x}
]
This simplifies to:
[
2.9 \times 10^{-8} = \frac{x^2}{0.265 – x}
]
Step 3: Make an approximation
Since Ka is very small, we can assume that the amount of dissociation (x) will be small compared to the initial concentration of HClO (0.265 M). Thus, ( 0.265 – x \approx 0.265 ). Substituting this into the equation:
[
2.9 \times 10^{-8} = \frac{x^2}{0.265}
]
Step 4: Solve for x
Now, solve for ( x ):
[
x^2 = (2.9 \times 10^{-8}) \times 0.265
]
[
x^2 = 7.685 \times 10^{-8}
]
[
x = \sqrt{7.685 \times 10^{-8}} = 8.77 \times 10^{-4} \text{ M}
]
Step 5: Calculate pH
The concentration of H+ is equal to ( x ), so:
[
[H^+] = 8.77 \times 10^{-4} \, \text{M}
]
The pH is given by:
[
\text{pH} = -\log[H^+] = -\log(8.77 \times 10^{-4}) = 3.06
]
Conclusion
The correct pH value is closest to 3.31. So, the correct answer is:
c) 3.31