Consider the reaction between strontium hydroxide and bromic acid, HBrO3(aq):
Sr(OH)2(aq) + 2 HBrO3(aq) →2 H2O (1) + Sr(BrO3)2(aq)
Calculate the concentration of bromic acid if a 20.95mL sample of bromic acid required 30.07mL of 0.383M Sr(OH)2 for complete neutralization to a phenolphthalein
endpoint.
The Correct Answer and Explanation is :
The concentration of bromic acid (HBrO3) is approximately 1.10 M.
Explanation:
To calculate the concentration of bromic acid, we use the stoichiometry of the reaction between strontium hydroxide (Sr(OH)2) and bromic acid (HBrO3):
[
\text{Sr(OH)}_2(aq) + 2 \text{HBrO}_3(aq) \rightarrow 2 \text{H}_2\text{O}(l) + \text{Sr(BrO}_3)_2(aq)
]
From the balanced equation, we know that 1 mole of Sr(OH)2 reacts with 2 moles of HBrO3. Using the volume and molarity of Sr(OH)2, we can calculate the moles of Sr(OH)2 that reacted.
- Calculate moles of Sr(OH)2:
Moles of Sr(OH)2 = Molarity × Volume (in L)
[
= 0.383 \, \text{M} \times \frac{30.07 \, \text{mL}}{1000} = 0.01152 \, \text{moles}
] - Use the reaction ratio to calculate moles of HBrO3:
From the equation, 1 mole of Sr(OH)2 reacts with 2 moles of HBrO3. Therefore, the moles of HBrO3 required will be:
[
0.01152 \, \text{moles of Sr(OH)2} \times 2 = 0.02304 \, \text{moles of HBrO3}
] - Calculate the concentration of HBrO3:
Molarity of HBrO3 = Moles of HBrO3 / Volume of HBrO3 (in L)
[
= \frac{0.02304 \, \text{moles}}{\frac{20.95 \, \text{mL}}{1000}} = 1.10 \, \text{M}
]
Thus, the concentration of bromic acid is approximately 1.10 M.
Now, I will generate an image summarizing these steps.
Here is the step-by-step diagram for the calculation of the bromic acid concentration, showing the stoichiometric relationship and the calculations involved. I hope it helps clarify the process!
