Answer:
0.1 liters of NaOH.
Explanation:
When the acid solution is neutralized we have:
[tex] n_{a} = n_{b} [/tex]
[tex] C_{a}V_{a} = C_{b}V_{b} [/tex]
Where:
[tex]n_{a}[/tex]: is the number of moles of the acid
[tex]n_{b}[/tex]: is the number of moles of the base
[tex]C_{a}[/tex]: is the concentration of the acid
[tex]C_{b}[/tex]: is the concentration of the base = 0.1 M
[tex]V_{a}[/tex]: is the volume of the acid = 10000 L
[tex]V_{b}[/tex]: is the volume of the base =?
The concentration of the acid can be calculated from the pH:
[tex] pH = -log([H^{+}]) [/tex]
[tex] [H^{+}] = 10^{-pH} = 10^{-6} M [/tex]
Now, we can find the volume of the base:
[tex] V_{b} = \frac{C_{a}V_{a}}{C_{b}} = \frac{10^{-6} M*10000 L}{0.1 M} = 0.1 L [/tex]
Therefore, the amount of NaOH needed to neutralize the solution is 0.1 liters.
I hope it helps you!
