Sure, let's tackle this problem step-by-step to find the pH when 45 mL of 0.8 M NaOH is mixed with 25 mL of 2.0 M HBr.
1. Calculate the number of moles of NaOH:
- Given the volume of NaOH solution: 45 mL = 0.045 L
- Molarity (M) of NaOH: 0.8 M
- Moles of NaOH = Molarity × Volume
[tex]\[
\text{Moles of NaOH} = 0.8 \, M \times 0.045 \, L = 0.036 \, \text{moles}
\][/tex]
2. Calculate the number of moles of HBr:
- Given the volume of HBr solution: 25 mL = 0.025 L
- Molarity (M) of HBr: 2.0 M
- Moles of HBr = Molarity × Volume
[tex]\[
\text{Moles of HBr} = 2.0 \, M \times 0.025 \, L = 0.05 \, \text{moles}
\][/tex]
3. Determine the moles of the excess reactant:
- Since HBr is a strong acid and NaOH is a strong base, they will neutralize each other in a 1:1 molar ratio.
- Calculate the difference in moles between HBr and NaOH:
[tex]\[
\text{Moles of excess reactant} = \text{Moles of HBr} - \text{Moles of NaOH}
\][/tex]
[tex]\[
\text{Moles of excess HBr} = 0.05 - 0.036 = 0.014 \, \text{moles}
\][/tex]
4. Calculate the final volume of the solution:
- The total volume is the sum of the volumes of the two solutions:
[tex]\[
\text{Final volume} = 45 \, \text{mL} + 25 \, \text{mL} = 70 \, \text{mL} = 0.07 \, \text{L}
\][/tex]
5. Calculate the concentration of the excess HBr:
- Concentration of HBr (which provides [tex]\( \text{H}_3\text{O}^+ \)[/tex]):
[tex]\[
\text{Concentration of } \text{H}_3\text{O}^+ = \frac{0.014 \, \text{moles}}{0.07 \, \text{L}} = 0.2 \, M
\][/tex]
6. Calculate the pH of the solution:
- pH is given by the negative logarithm of the hydrogen ion concentration:
[tex]\[
\text{pH} = -\log_{10}([ \text{H}_3\text{O}^+])
\][/tex]
[tex]\[
\text{pH} = -\log_{10}(0.2) \approx 0.70
\][/tex]
So, the pH of the solution when 45 mL of 0.8 M NaOH is mixed with 25 mL of 2.0 M HBr is approximately 0.70.
