15 mL of 0.1M NaOH was used to titrate 30 ml of HI solution until the end point was reached. What is the concentration of the HI?
A 0.10 M
B 0.05 M
C 0.40 M
D 0.20 M
The correct answer and explanation is :
To find the concentration of HI (hydroiodic acid) in this titration, we can use the concept of moles and stoichiometry. Here’s the step-by-step breakdown:
Given:
- Volume of NaOH = 15 mL = 0.015 L
- Concentration of NaOH = 0.1 M
- Volume of HI = 30 mL = 0.030 L
- Titration reaction: The balanced chemical equation for the titration of NaOH with HI is: [
NaOH + HI \rightarrow NaI + H_2O
]
This shows that NaOH reacts with HI in a 1:1 molar ratio. So, for every mole of NaOH, one mole of HI reacts.
Step 1: Calculate the moles of NaOH used
The number of moles of NaOH can be calculated using the formula:
[
\text{moles of NaOH} = M \times V
]
Where:
- ( M = 0.1 \, \text{M} ) (molarity of NaOH)
- ( V = 0.015 \, \text{L} ) (volume of NaOH in liters)
[
\text{moles of NaOH} = 0.1 \, \text{M} \times 0.015 \, \text{L} = 0.0015 \, \text{moles of NaOH}
]
Step 2: Use the stoichiometry of the reaction
From the balanced equation, we know that the mole ratio of NaOH to HI is 1:1. This means that the moles of HI that reacted are also 0.0015 moles.
Step 3: Calculate the concentration of HI
Now, we can calculate the concentration of HI using the formula:
[
\text{molarity of HI} = \frac{\text{moles of HI}}{\text{volume of HI in liters}}
]
[
\text{molarity of HI} = \frac{0.0015 \, \text{moles}}{0.030 \, \text{L}} = 0.05 \, \text{M}
]
Conclusion:
The concentration of HI is 0.05 M, so the correct answer is:
B) 0.05 M.
This calculation assumes complete neutralization, as indicated by the fact that the titration reached the endpoint, meaning all the HI has reacted with NaOH.