Question

250 mL 0.1 M HCl solution is mixed with 250 mL

0.2 M NaOH. The concentration of OH- in the
mixture is
​

Answer 1

The concentration of OH⁻ in the mixture is 0.05 M

Step-by-step 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 =(n_(NaOH))/(V_(T))=(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!