# How many millilitres of 0.200 m naoh are required to neutralize 20.0 ml of 0.100 m hcl?

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The question is –

## How many millilitres of 0.200 m NaOH are required to neutralize 20.0 ml of 0.100 m HCl?

⇒ 10 ml of 0.200 M NaOH is required to neutralize 20.0 ml of 0.100 M HCl.

Explanation:

We can solve this question using two methods, Let’s start with Method 1.

To neutralize an acid with a base, you need to use an equal amount of base as the acid. The acid and base will react to form water and salt. The balanced equation for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is:

HCl + NaOH → NaCl + H2O

To find out how many milliliters of 0.200 M NaOH is required to neutralize 20.0 ml of 0.100 M HCl, we can use the following formula:

∴ Molarity = moles of solute/liters of solution

We can use this formula to find the moles of HCl and NaOH present in the solutions.

⇒ Moles HCl = (Molarity HCl) * (Volume HCl) = (0.100 M) * (20.0 ml) = 2.00 x 10-3 moles

⇒ Moles NaOH = (Molarity NaOH) * (Volume NaOH) = (0.200 M) *(x)  = 2.00 x 10-3 moles

We can use this information to set up a proportion to find the volume of NaOH required.

⇒ (0.200 M) * (x) = 2.00 x 10-3 moles

∴ x = 2.00 x 10-3 moles/(0.200 M) = 0.01 L = 10 ml

So, 10 ml of 0.200 M NaOH is required to neutralize 20.0 ml of 0.100 M HCl.

Method 2:

We can use the formula, M1V1 = M2V2 to find the volume of base (NaOH) required to neutralize a certain volume of acid (HCl).

To use this formula, we need to know the molarity (m) and volume (v) of both the acid and the base.

In this case, we know the following:

• molarity of HCl = 0.100 M
• volume of HCl = 20.0 ml
• molarity of NaOH = 0.200 M

We want to find the volume of NaOH (V2) required to neutralize 20.0 ml of 0.100 M HCl.

The equation M1V1 = M2V2 can be rearranged to solve for V2:

V2 = M1V1/M2

⇒ We can plug in the known values to find the volume of NaOH required:

∴  V2 = (0.100 M) * (20.0 ml) / (0.200 M) = 10 ml

So, 10 ml of 0.200 M NaOH is required to neutralize 20.0 ml of 0.100 M HCl.

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