How many milliliters of 0.135 M HCl are needed to completely neutralize 56.0 mL of 0.101 M Ba(OH)2 solution?
You need an equation.
Ba(OH)2 + 2HCl --> BaCl2 + 2H2O
Think moles.'
moles Ba(OH)2 = M x L = ??
Using the coefficients in the balanced equation, convert moles Ba(OH)2 to moles HCl.
??moles Ba(OH)2 x (2 moles HCl/1 mole Ba(OH)2) = xx moles HCl
Tben the definition of molarity is
M = moles/L.
You have moles HCl and M HCl, solve for L HCl and convert to mL.
To determine how many milliliters of the HCl solution are needed to neutralize the Ba(OH)2 solution, we can use the concept of stoichiometry, which relates the moles of one substance to another in a balanced chemical equation.
First, let's write the balanced chemical equation for the reaction between HCl and Ba(OH)2:
2 HCl + Ba(OH)2 -> 2 H2O + BaCl2
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Ba(OH)2 to produce 2 moles of H2O and 1 mole of BaCl2.
Next, we need to determine the number of moles of Ba(OH)2 in the given 56.0 mL solution. We can use the formula:
moles = concentration x volume
moles of Ba(OH)2 = 0.101 M x 0.0560 L = 0.005656 moles
Since the stoichiometry of the balanced equation tells us that 2 moles of HCl react with 1 mole of Ba(OH)2, we can determine the number of moles of HCl needed to neutralize the Ba(OH)2 solution:
moles of HCl = (1/2) x moles of Ba(OH)2 = (1/2) x 0.005656 moles = 0.002828 moles
Finally, we can determine the volume of the 0.135 M HCl solution needed to contain 0.002828 moles of HCl using the formula:
volume = moles / concentration
volume = 0.002828 moles / 0.135 M = 0.02097 liters
To convert this volume to milliliters, we multiply by 1000:
volume = 0.02097 liters x 1000 = 20.97 mL
Therefore, approximately 20.97 milliliters of the 0.135 M HCl solution are needed to completely neutralize the 56.0 mL of the 0.101 M Ba(OH)2 solution.