What is the volume of 0.05-molar HCl that is required to neutralize 50 ml of a 0.10-molar Mg(OH)2 solution?
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
mols Mg(OH)2 = M x L = ??
moles HCl = 2x that (Look at the coefficients in the balanced equation.)
M HCl = mols HCl/L HCl
Substitute and solve for L HCl.
To determine the volume of 0.05-molar HCl required to neutralize 50 ml of a 0.10-molar Mg(OH)2 solution, we can use the concept of stoichiometry. The balanced equation for the reaction between HCl and Mg(OH)2 is:
2HCl + Mg(OH)2 → MgCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2. In other words, the stoichiometric ratio between HCl and Mg(OH)2 is 2:1.
Step 1: Calculate the number of moles of Mg(OH)2 in 50 ml of the 0.10-molar solution.
Molarity (M) = moles/volume (L)
0.10 M = moles/0.050 L
moles = 0.10 M x 0.050 L
moles = 0.005 moles
Step 2: Use the stoichiometric ratio to determine the number of moles of HCl required.
From the equation, we know that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
moles HCl = 2 x moles Mg(OH)2
moles HCl = 2 x 0.005 moles
moles HCl = 0.01 moles
Step 3: Calculate the volume of 0.05-molar HCl solution containing 0.01 moles.
Molarity (M) = moles/volume (L)
0.05 M = 0.01 moles/volume (L)
volume (L) = 0.01 moles / 0.05 M
volume (L) = 0.2 L
Step 4: Convert the volume from liters to milliliters.
1 L = 1000 ml
volume (ml) = 0.2 L x 1000 ml/L
volume (ml) = 200 ml
Therefore, the volume of 0.05-molar HCl required to neutralize 50 ml of a 0.10-molar Mg(OH)2 solution is 200 ml.
To find the volume of 0.05-molar HCl needed to neutralize the 0.10-molar Mg(OH)2 solution, we can use the concept of stoichiometry.
Stoichiometry involves the balanced chemical equation of the reaction between HCl and Mg(OH)2 to determine the mole ratio between the two reactants. The balanced equation for the reaction is:
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2.
To solve the problem, follow these steps:
Step 1: Convert the given volume of Mg(OH)2 solution (50 ml) to moles.
To do this, we need to use the molarity (0.10 M) and the formula:
moles = volume (in liters) x molarity
Given:
Volume of Mg(OH)2 = 50 ml = 50/1000 = 0.05 L
Molarity of Mg(OH)2 = 0.10 M
moles of Mg(OH)2 = 0.05 L x 0.10 M = 0.005 moles
Step 2: Determine the mole ratio between HCl and Mg(OH)2
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2. This means that for every 1 mole of Mg(OH)2, we need 2 moles of HCl.
Step 3: Calculate the volume of HCl needed.
Use the mole ratio to calculate the moles of HCl required to neutralize the given amount of Mg(OH)2.
moles of HCl = moles of Mg(OH)2 x (2 moles of HCL / 1 mole of Mg(OH)2)
moles of HCl = 0.005 moles x (2 / 1) = 0.01 moles
Step 4: Convert moles of HCl to volume.
Use the molarity of HCl (0.05 M) to convert the moles of HCl to volume in liters.
volume of HCl = moles of HCl / molarity of HCl
volume of HCl = 0.01 moles / 0.05 M = 0.2 L
Step 5: Convert the volume to milliliters (ml).
volume of HCl = 0.2 L x 1000 ml/L = 200 ml
Therefore, you would need 200 ml of 0.05-molar HCl to neutralize 50 ml of a 0.10-molar Mg(OH)2 solution.