Question

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

Answer 1

Answer:

0.1 liters of NaOH.

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:

$V_{b} = \frac{C_{a}V_{a}}{C_{b}} = \frac{10^{-6} M*10000 L}{0.1 M} = 0.1 L$

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

I hope it helps you!