a volume of 50mL of a standard HCl solution is needed to neutralize 50 mL of 1x10-2 M NaOH. what volume of this HCl solution is needed to neutralize 50 mL of 1x10-2 M Ba(OH)2
First the M of HCl must be determined from the titration with NaOH.
NaOH + HCl ==> NaCl + H2O
Since the molar ratio or the reactants is 1:1, one may use
mL x M = mL x M. Use that to determine M HCl.
Then go to the Ca(OH)2 solution.
2HCl + Ba(OH)2 ==> BaCl2 + 2H2O
Here the ratio of the reactants is not 1:1; therefore, a different approach must be used.
moles Ba(OH)2 = M x L.
Moles HCl = twice that (from the coefficients in the balanced equation).
moles HCl = M x L
You know moles and M, calculat L and convert to mL.
To determine the volume of the HCl solution needed to neutralize 50 mL of 1x10^-2 M Ba(OH)2, we can use the concept of stoichiometry and the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction between HCl and Ba(OH)2 is:
2 HCl + Ba(OH)2 → BaCl2 + 2 H2O
From the equation, we can see that 2 moles of HCl are required to react with 1 mole of Ba(OH)2.
Given that the volume of the HCl solution needed to neutralize 50 mL of 1x10^-2 M NaOH is also 50 mL, we can use this information to determine the number of moles of HCl required to neutralize 1x10^-2 M of NaOH.
The molarity (M) of a solution is defined as the number of moles of solute per liter of solution. Thus, the number of moles of HCl in the 50 mL solution can be calculated as:
Moles of HCl = Molarity × Volume (in liters)
= (1x10^-2 M) × (50 mL / 1000 mL/L)
= 5x10^-4 moles of HCl
Now, we can use the stoichiometric ratio from the balanced equation to determine the number of moles of Ba(OH)2 that can be neutralized by the 5x10^-4 moles of HCl:
Moles of Ba(OH)2 = (Moles of HCl / 2)
= (5x10^-4 moles / 2)
= 2.5x10^-4 moles of Ba(OH)2
Finally, we need to calculate the volume of the HCl solution required to neutralize 2.5x10^-4 moles of Ba(OH)2. Using the given molarity (1x10^-2 M) of the Ba(OH)2 solution and the volume-molarity relationship, we can solve for the volume:
Volume (in liters) = Moles / Molarity
= (2.5x10^-4 moles) / (1x10^-2 M)
= 2.5x10^-2 L
Converting the volume from liters to milliliters:
Volume (in mL) = (2.5x10^-2 L) × (1000 mL / 1 L)
= 25 mL
Therefore, 25 mL of the HCl solution is needed to neutralize 50 mL of 1x10^-2 M Ba(OH)2.