Asked by Samantha simbi
What volume of 0.075M HNO3 is required to neutralise 100mL of 0.01M CA(OH)2 solution.
Answers
Answered by
GPT 3.5
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).
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).