How many mL of a 0.5M Ca(OH)2 solution are needed to neutralize 38mL of a 2M HNO3 solution?
2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O
mols HNO3 = M x L = ?
Convert mols HNO3 to mols Ca(OH)2 using the coefficients in the balanced equation.
The mols Ca(OH)2 = M x L. You know M and mols, solve for L and convert to mL.
To find out how many mL of the 0.5M Ca(OH)2 solution are needed to neutralize the 38mL of the 2M HNO3 solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction.
The balanced chemical equation for the reaction between Ca(OH)2 and HNO3 is:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
From the equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HNO3. In other words, the stoichiometric ratio between Ca(OH)2 and HNO3 is 1:2.
Now, let's calculate the number of moles of HNO3 in 38mL of the 2M HNO3 solution:
Number of moles = Concentration (M) × Volume (L)
= 2M × 0.038L (since 1 mL = 0.001 L)
= 0.076 moles
According to the stoichiometry, 1 mole of Ca(OH)2 is required to neutralize 2 moles of HNO3. So, the number of moles of Ca(OH)2 needed to neutralize the HNO3 is also 0.076 moles.
Now, let's calculate the volume of the 0.5M Ca(OH)2 solution needed to provide 0.076 moles of Ca(OH)2:
Volume (L) = Number of moles / Concentration (M)
= 0.076 moles / 0.5M
= 0.152 L
Finally, convert the volume to milliliters:
Volume (mL) = Volume (L) × 1000
= 0.152 L × 1000
= 152 mL
Therefore, 152 mL of the 0.5M Ca(OH)2 solution are needed to neutralize 38 mL of the 2M HNO3 solution.