Calculate the pH when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO (Ka = 2.5 × 10��)
The Correct Answer and Explanation is :
To calculate the pH of the solution when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO, we need to follow these steps:
Step 1: Determine the moles of KOH and HBrO
- Moles of KOH:
[
\text{Moles of KOH} = Molarity \times Volume = 0.150 \, \text{M} \times 0.020 \, \text{L} = 0.0030 \, \text{mol}
] - Moles of HBrO:
[
\text{Moles of HBrO} = Molarity \times Volume = 0.300 \, \text{M} \times 0.038 \, \text{L} = 0.0114 \, \text{mol}
]
Step 2: Reaction between KOH and HBrO
KOH is a strong base, and HBrO is a weak acid. The neutralization reaction between KOH and HBrO is:
[
\text{KOH} + \text{HBrO} \rightarrow \text{KBrO} + \text{H}_2\text{O}
]
Since there are fewer moles of KOH than HBrO, KOH will neutralize an equivalent amount of HBrO. After the reaction, the remaining HBrO will determine the pH.
- Moles of HBrO remaining = (0.0114 – 0.0030 = 0.0084 \, \text{mol})
Step 3: Concentration of remaining HBrO
The total volume of the mixture is:
[
\text{Volume of solution} = 20.0 \, \text{mL} + 38.0 \, \text{mL} = 58.0 \, \text{mL} = 0.0580 \, \text{L}
]
The concentration of remaining HBrO is:
[
[\text{HBrO}] = \frac{0.0084 \, \text{mol}}{0.0580 \, \text{L}} = 0.1448 \, \text{M}
]
Step 4: Calculate the pH
HBrO partially dissociates in water, and the dissociation equilibrium is:
[
\text{HBrO} \rightleftharpoons \text{H}^+ + \text{BrO}^-
]
The acid dissociation constant ((K_a)) is given as (2.5 \times 10^{-9}). We can set up an equilibrium expression:
[
K_a = \frac{[\text{H}^+][\text{BrO}^-]}{[\text{HBrO}]}
]
Assume that the concentration of ( \text{H}^+ ) and ( \text{BrO}^- ) formed at equilibrium is (x), and the concentration of HBrO at equilibrium is (0.1448 – x). The equation becomes:
[
K_a = \frac{x^2}{0.1448 – x}
]
Given that (K_a) is very small, we can approximate (0.1448 – x \approx 0.1448). Thus, the equation simplifies to:
[
2.5 \times 10^{-9} = \frac{x^2}{0.1448}
]
Solving for (x):
[
x^2 = (2.5 \times 10^{-9})(0.1448) = 3.62 \times 10^{-10}
]
[
x = \sqrt{3.62 \times 10^{-10}} = 6.02 \times 10^{-5} \, \text{M}
]
Thus, the concentration of ( \text{H}^+ ) is (6.02 \times 10^{-5} \, \text{M}). To find the pH:
[
\text{pH} = -\log [\text{H}^+] = -\log (6.02 \times 10^{-5}) = 4.22
]
Conclusion:
The pH of the solution after mixing 20.0 mL of 0.150 M KOH with 38.0 mL of 0.300 M HBrO is 4.22.
Let me now generate a relevant image for this calculation.
Here is an image that visually illustrates the neutralization between KOH and HBrO, showing the dissociation and pH calculation steps. Let me know if you need further clarification!
