Question

Calculate the amount of 0.1 M base needed to neutralize 10,000 liters of pH 6.0 water.

The base is Sodium Hydroxide (NaOH) Please show work!

Answer 1

0.1 liters of NaOH.

Step-by-step explanation:

When the acid solution is neutralized we have:

`n_(a) = n_(b)`

`C_(a)V_(a) = C_(b)V_(b)`

Where:

`n_(a)`: is the number of moles of the acid

`n_(b)`: is the number of moles of the base

`C_(a)`: is the concentration of the acid

`C_(b)`: is the concentration of the base = 0.1 M

`V_(a)`: is the volume of the acid = 10000 L

`V_(b)`: is the volume of the base =?

The concentration of the acid can be calculated from the pH:

`pH = -log([H^(+)])`

`[H^(+)] = 10^(-pH) = 10^(-6) M`

Now, we can find the volume of the base:

`V_(b) = (C_(a)V_(a))/(C_(b)) = (10^(-6) M*10000 L)/(0.1 M) = 0.1 L`

Therefore, the amount of NaOH needed to neutralize the solution is 0.1 liters.

I hope it helps you!