Calculate the volume of 0.100 mol of dm^-3 nitric acid needed to react with 20ml of mol dm-^3 of 0.120 mol dm-^3sodium hydroxide ?

The balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH) is:

HNO3 + NaOH -> NaNO3 + H2O

From the equation, we can see that 1 mole of nitric acid reacts with 1 mole of sodium hydroxide.

Given that the concentration of sodium hydroxide is 0.120 mol dm^-3 and the volume is 20 ml (which is equal to 0.020 dm^3), we can calculate the number of moles of sodium hydroxide:

moles of NaOH = concentration x volume
moles of NaOH = 0.120 mol dm^-3 x 0.020 dm^3
moles of NaOH = 0.0024 moles

Since the reaction is 1:1, we need the same number of moles of nitric acid to react with the sodium hydroxide. Therefore, we need 0.0024 moles of nitric acid.

Now, we are given that the concentration of nitric acid is 0.100 mol dm^-3. We can calculate the volume of nitric acid needed:

volume of HNO3 = moles / concentration
volume of HNO3 = 0.0024 moles / 0.100 mol dm^-3
volume of HNO3 = 0.024 dm^3

Therefore, 0.024 dm^3 (or 24 ml) of 0.100 mol dm^-3 nitric acid is needed to react with 20 ml of 0.120 mol dm^-3 sodium hydroxide.

First, write the balanced chemical equation for the reaction between nitric acid and sodium hydroxide:

```
HNO3 + NaOH -> NaNO3 + H2O
```

From the balanced equation, we see that 1 mol of nitric acid reacts with 1 mol of sodium hydroxide.

To find the volume of 0.100 mol dm^-3 nitric acid needed, we can use the formula:

```
Volume = moles of solute / concentration
```

where:

- Volume is in liters
- moles of solute is in moles
- concentration is in mol/L

Substituting the given values into the formula, we get:

```
Volume = 0.120 mol / 0.100 mol/dm^-3 = 1.20 dm^3
```

Therefore, we need 1.20 dm^3 of 0.100 mol dm^-3 nitric acid to react with 20ml of mol dm-^3 of 0.120 mol dm-^3sodium hydroxide.