Answer :
Sure, let's evaluate the pH of the solution when 45 mL of 0.8 M NaOH is mixed with 18 mL of 2 M HBr. We'll go through this step-by-step.
### Step 1: Calculate the Moles of NaOH and HBr
1. Moles of NaOH:
- Volume of NaOH solution: [tex]\(45\)[/tex] mL [tex]\((45 \times 10^{-3}\)[/tex] L)
- Concentration of NaOH solution: [tex]\(0.8\)[/tex] M
Using the formula [tex]\( \text{Moles} = \text{Molarity} \times \text{Volume} \)[/tex]:
[tex]\[ \text{Moles of NaOH} = 0.8 \, \text{M} \times 45 \times 10^{-3} \, \text{L} = 0.036 \, \text{mol} \][/tex]
2. Moles of HBr:
- Volume of HBr solution: [tex]\(18\)[/tex] mL [tex]\((18 \times 10^{-3}\)[/tex] L)
- Concentration of HBr solution: [tex]\(2.0\)[/tex] M
Using the formula [tex]\( \text{Moles} = \text{Molarity} \times \text{Volume} \)[/tex]:
[tex]\[ \text{Moles of HBr} = 2.0 \, \text{M} \times 18 \times 10^{-3} \, \text{L} = 0.036 \, \text{mol} \][/tex]
### Step 2: Determine the Limiting Reagent and Net Moles of [tex]\( \text{H}^+ \)[/tex] or [tex]\( \text{OH}^- \)[/tex]
NaOH is a strong base and dissociates completely in water to give [tex]\( \text{OH}^- \)[/tex] ions.
HBr is a strong acid and dissociates completely in water to give [tex]\( \text{H}^+ \)[/tex] ions.
The reaction between NaOH and HBr is:
[tex]\[ \text{NaOH} + \text{HBr} \rightarrow \text{NaBr} + \text{H}_2\text{O} \][/tex]
From the reaction, 1 mole of NaOH reacts with 1 mole of HBr. Given that the moles of NaOH and HBr both equal [tex]\(0.036\)[/tex] mol, they will completely neutralize each other.
### Step 3: Determine the Final Species in Solution
Since the moles of NaOH and HBr are equal, they completely neutralize each other, leaving no excess [tex]\( \text{H}^+ \)[/tex] or [tex]\( \text{OH}^- \)[/tex] ions in the solution. The solution is now neutral with a pH of 7.
However, let's confirm the total volume of the mixture:
[tex]\[ \text{Total Volume} = 45 \, \text{mL} + 18 \, \text{mL} = 63 \, \text{mL} \, (0.063 \, \text{L}) \][/tex]
### Conclusion
The reaction between 45 mL of 0.8 M NaOH and 18 mL of 2 M HBr results in a neutral solution with no excess ions. Therefore, the pH of the resulting solution is 7.
So, the final answer is:
[tex]\[ \boxed{7} \][/tex]
### Step 1: Calculate the Moles of NaOH and HBr
1. Moles of NaOH:
- Volume of NaOH solution: [tex]\(45\)[/tex] mL [tex]\((45 \times 10^{-3}\)[/tex] L)
- Concentration of NaOH solution: [tex]\(0.8\)[/tex] M
Using the formula [tex]\( \text{Moles} = \text{Molarity} \times \text{Volume} \)[/tex]:
[tex]\[ \text{Moles of NaOH} = 0.8 \, \text{M} \times 45 \times 10^{-3} \, \text{L} = 0.036 \, \text{mol} \][/tex]
2. Moles of HBr:
- Volume of HBr solution: [tex]\(18\)[/tex] mL [tex]\((18 \times 10^{-3}\)[/tex] L)
- Concentration of HBr solution: [tex]\(2.0\)[/tex] M
Using the formula [tex]\( \text{Moles} = \text{Molarity} \times \text{Volume} \)[/tex]:
[tex]\[ \text{Moles of HBr} = 2.0 \, \text{M} \times 18 \times 10^{-3} \, \text{L} = 0.036 \, \text{mol} \][/tex]
### Step 2: Determine the Limiting Reagent and Net Moles of [tex]\( \text{H}^+ \)[/tex] or [tex]\( \text{OH}^- \)[/tex]
NaOH is a strong base and dissociates completely in water to give [tex]\( \text{OH}^- \)[/tex] ions.
HBr is a strong acid and dissociates completely in water to give [tex]\( \text{H}^+ \)[/tex] ions.
The reaction between NaOH and HBr is:
[tex]\[ \text{NaOH} + \text{HBr} \rightarrow \text{NaBr} + \text{H}_2\text{O} \][/tex]
From the reaction, 1 mole of NaOH reacts with 1 mole of HBr. Given that the moles of NaOH and HBr both equal [tex]\(0.036\)[/tex] mol, they will completely neutralize each other.
### Step 3: Determine the Final Species in Solution
Since the moles of NaOH and HBr are equal, they completely neutralize each other, leaving no excess [tex]\( \text{H}^+ \)[/tex] or [tex]\( \text{OH}^- \)[/tex] ions in the solution. The solution is now neutral with a pH of 7.
However, let's confirm the total volume of the mixture:
[tex]\[ \text{Total Volume} = 45 \, \text{mL} + 18 \, \text{mL} = 63 \, \text{mL} \, (0.063 \, \text{L}) \][/tex]
### Conclusion
The reaction between 45 mL of 0.8 M NaOH and 18 mL of 2 M HBr results in a neutral solution with no excess ions. Therefore, the pH of the resulting solution is 7.
So, the final answer is:
[tex]\[ \boxed{7} \][/tex]
