Question

250 mL 0.1 M HCl solution is mixed with 250 mL

Answer 1

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:

$n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles$

Similarly, the number of moles of NaOH is:

$n_{NaOH} = C*V = 0.2 mol/L*0.250L = 0.05 moles$

Now, from the reaction of HCl and NaOH we have the following number of moles of NaOH remaining:

$n_{NaOH} = 0.05 moles - 0.025 moles = 0.025 moles$

Finally, the concentration of OH⁻ in the mixture is:

$C =\frac{n_{NaOH}}{V_{T}}=\frac{0.025 moles}{0.250*2 L} = 0.05 moles/L$

Therefore, the concentration of OH⁻ in the mixture is 0.05 M.

I hope it helps you!