How many milliliters of 0.155 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution?
Do this the same way you solved the iron(II)/NaOH problem earlier.
To find out how many milliliters of 0.155 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Ba(OH)2.
The balanced chemical equation for the reaction is:
2HCl + Ba(OH)2 → BaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2.
Step 1: Convert the given volume of Ba(OH)2 solution to moles.
0.101 M represents 0.101 moles of Ba(OH)2 in 1 liter of the solution.
So, in 35.0 mL (0.035 liters) of the solution, the number of moles of Ba(OH)2 is:
0.101 moles/L * 0.035 L = 0.003535 moles.
Step 2: Determine the number of moles of HCl required using stoichiometry.
According to the balanced equation, 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2.
Therefore, the number of moles of HCl required is:
2 * 0.003535 moles = 0.00707 moles.
Step 3: Convert the moles of HCl to milliliters of the 0.155 M HCl solution.
0.155 M represents 0.155 moles of HCl in 1 liter of the solution.
So, to calculate the volume of 0.155 M HCl solution required, divide the number of moles by the molarity:
0.00707 moles / 0.155 moles/L = 0.0457 L.
Since the given molarity is in liters, convert it to milliliters by multiplying by 1000:
0.0457 L * 1000 mL/L = 45.7 mL.
Therefore, approximately 45.7 milliliters of 0.155 M HCl are needed to completely neutralize 35.0 mL of 0.101 M Ba(OH)2 solution.
To find the amount of HCl needed to neutralize the Ba(OH)2 solution, we can use the equation:
moles HCl = moles Ba(OH)2
First, calculate the moles of Ba(OH)2:
moles Ba(OH)2 = volume (in liters) x molarity
= (35.0 mL / 1000 mL/L) x 0.101 mol/L
= 0.035 L x 0.101 mol/L
= 0.003535 mol
Since the reaction is 1:2 between Ba(OH)2 and HCl (1 mole of Ba(OH)2 reacts with 2 moles of HCl), we need twice the number of moles of HCl:
moles HCl = 2 x moles Ba(OH)2
= 2 x 0.003535 mol
= 0.00707 mol
Finally, we can calculate the volume of the 0.155 M HCl solution needed:
volume HCl = moles HCl / molarity
= 0.00707 mol / 0.155 mol/L
= 0.0457 L
Convert the volume to milliliters:
volume HCl = 0.0457 L x 1000 mL/L
= 45.7 mL
Therefore, 45.7 milliliters of 0.155 M HCl are needed to neutralize completely 35.0 mL of 0.101 M Ba(OH)2 solution.