What volume of 0.075M HNO3 is required to neutralise 100mL of 0.01M CA(OH)2 solution.

1 answer

The balanced chemical equation for the neutralization reaction between HNO3 and Ca(OH)2 is:

2 HNO3 + Ca(OH)2 -> Ca(NO3)2 + 2 H2O

From the balanced equation, we can see that the stoichiometric ratio between HNO3 and Ca(OH)2 is 2:1.

Given that the concentration of Ca(OH)2 is 0.01 M and the volume is 100 mL, we can calculate the number of moles of Ca(OH)2:

Moles of Ca(OH)2 = concentration * volume
= 0.01 M * 0.1 L
= 0.001 moles

Since the stoichiometric ratio between HNO3 and Ca(OH)2 is 2:1, the number of moles of HNO3 required to neutralize the Ca(OH)2 is the same:

Moles of HNO3 = 0.001 moles

To find the volume of 0.075 M HNO3 needed to neutralize the Ca(OH)2, we can use the equation:

Moles = concentration * volume

So, we rearrange the equation to solve for volume:

Volume = Moles / concentration

Plugging in the values:

Volume = 0.001 moles / 0.075 M
Volume = 0.01333 L

Therefore, the volume of 0.075 M HNO3 required to neutralize 100 mL of 0.01 M Ca(OH)2 solution is 0.01333 L or 13.33 mL (rounded to 2 decimal places).