What volume of 0.075M HNO3 is required to neutralise 100mL of 0.01M CA(OH)2 solution.
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).
To determine the volume of 0.075M HNO3 solution required to neutralize 100mL of 0.01M Ca(OH)2 solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HNO3 and Ca(OH)2.
The balanced chemical equation for the reaction is:
2 HNO3 + Ca(OH)2 -> Ca(NO3)2 + 2 H2O
From the equation, we can see that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
First, let's calculate the number of moles of Ca(OH)2 in 100mL of 0.01M solution:
Moles of Ca(OH)2 = Volume (L) x Concentration (M)
= 0.1 L x 0.01 M
= 0.001 moles
Since the ratio of HNO3 to Ca(OH)2 is 2:1, we would need twice the number of moles of HNO3 to neutralize the Ca(OH)2. Therefore:
Moles of HNO3 = 2 x Moles of Ca(OH)2
= 2 x 0.001 moles
= 0.002 moles
To calculate the required volume of 0.075M HNO3 solution, we can use the formula:
Volume (L) = Moles / Concentration (M)
= 0.002 moles / 0.075 M
Calculating the volume:
Volume (L) = 0.0267 L
Finally, we convert the volume to milliliters:
Volume (mL) = Volume (L) x 1000 mL/L
= 0.0267 L x 1000 mL/L
= 26.7 mL
Therefore, approximately 26.7 mL of 0.075M HNO3 solution is required to neutralize 100mL of 0.01M Ca(OH)2 solution.