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RUDIKE [14]
3 years ago
9

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

Chemistry
1 answer:
musickatia [10]3 years ago
8 0

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!    

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