Answer:
The concentration of OH⁻ in the mixture is 0.05 M
Explanation:
The reaction of neutralization between HCl and NaOH is the following:
H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)
The number of moles of HCl is:
[tex] n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles [/tex]
Similarly, the number of moles of NaOH is:
[tex] n_{NaOH} = C*V = 0.2 mol/L*0.250L = 0.05 moles [/tex]
Now, from the reaction of HCl and NaOH we have the following number of moles of NaOH remaining:
[tex] n_{NaOH} = 0.05 moles - 0.025 moles = 0.025 moles [/tex]
Finally, the concentration of OH⁻ in the mixture is:
[tex]C =\frac{n_{NaOH}}{V_{T}}=\frac{0.025 moles}{0.250*2 L} = 0.05 moles/L[/tex]
Therefore, the concentration of OH⁻ in the mixture is 0.05 M.
I hope it helps you!
