Given the balanced equation: HBr + NaOH -> NaBr + H2O (a) Calculate the number of moles of HBr in 100.0 mL of 0.250 M HBr. Calculate the number of moles of NaOH in 100.0 mL of 0.250 M NaOH. When these two solutions are mixed, the acid and base should neutralize each other exactly. This means that all of the acid and base are completely used up; either one could be considered the “limiting reactant”. Starting with the moles of either the acid or base, calculate the number of moles of salt produced by the reaction. Using the moles of salt produced and the total volume in mL of the solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction:
The Correct Answer and Explanation is:
Step-by-Step Solution:
Given the reaction:
HBr+NaOH→NaBr+H2O\text{HBr} + \text{NaOH} \rightarrow \text{NaBr} + \text{H}_2\text{O}HBr+NaOH→NaBr+H2O
This is a neutralization reaction where HBr (hydrobromic acid) reacts with NaOH (sodium hydroxide) to form NaBr (sodium bromide) and water.
1. Calculate moles of HBr:
We are given the molarity (M) and volume (V) of the HBr solution.
Molarity (M) = 0.250 M
Volume (V) = 100.0 mL = 0.100 L (converted from mL to L)
To calculate the number of moles, we use the formula:moles of HBr=Molarity×Volume\text{moles of HBr} = \text{Molarity} \times \text{Volume}moles of HBr=Molarity×Volumemoles of HBr=0.250 M×0.100 L=0.0250 moles\text{moles of HBr} = 0.250 \, \text{M} \times 0.100 \, \text{L} = 0.0250 \, \text{moles}moles of HBr=0.250M×0.100L=0.0250moles
2. Calculate moles of NaOH:
Similarly, for NaOH, we are also given:
Molarity (M) = 0.250 M
Volume (V) = 100.0 mL = 0.100 L
Using the same formula:moles of NaOH=Molarity×Volume\text{moles of NaOH} = \text{Molarity} \times \text{Volume}moles of NaOH=Molarity×Volumemoles of NaOH=0.250 M×0.100 L=0.0250 moles\text{moles of NaOH} = 0.250 \, \text{M} \times 0.100 \, \text{L} = 0.0250 \, \text{moles}moles of NaOH=0.250M×0.100L=0.0250moles
3. Determine moles of salt (NaBr) produced:
From the balanced equation, we can see that for every mole of HBr that reacts, one mole of NaOH is consumed, and one mole of NaBr is produced.
Since we have equal moles of HBr and NaOH (0.0250 moles each), both will be completely consumed, and the number of moles of NaBr produced is also 0.0250 moles.
4. Calculate the molarity of NaBr solution:
After the reaction, the total volume of the solution will be the sum of the volumes of HBr and NaOH solutions:Total Volume=100.0 mL+100.0 mL=200.0 mL=0.200 L\text{Total Volume} = 100.0 \, \text{mL} + 100.0 \, \text{mL} = 200.0 \, \text{mL} = 0.200 \, \text{L}Total Volume=100.0mL+100.0mL=200.0mL=0.200L
Now, we can calculate the molarity of the NaBr solution using the formula:Molarity of NaBr=moles of NaBrtotal volume in L\text{Molarity of NaBr} = \frac{\text{moles of NaBr}}{\text{total volume in L}}Molarity of NaBr=total volume in Lmoles of NaBrMolarity of NaBr=0.0250 moles0.200 L=0.125 M\text{Molarity of NaBr} = \frac{0.0250 \, \text{moles}}{0.200 \, \text{L}} = 0.125 \, \text{M}Molarity of NaBr=0.200L0.0250moles=0.125M
Final Answer:
- Moles of HBr = 0.0250 moles
- Moles of NaOH = 0.0250 moles
- Moles of NaBr produced = 0.0250 moles
- Molarity of NaBr solution = 0.125 M
