1. (a) What is the molarity of a solution prepared by dissolving 4.315 g of NaOH in water to exactly 100 mL in a graduated flask? (Na = 23.0; O = 16.0' H = 1.0 amu).
(b) What volume of the above solution is required to neutralize exactly 25.00 mL of 0.1500 M HCl solution?
NaOH(aq) + HCl(aq) = NaCl(aq) + H2O(l)
mol NaOH = grams/molar mass= ?
M = mols/L = ? mols/0.1L
b)
mols HCl = M x L = ?
mols NaOH = mols HCl
M NaOH = mols/L soln. You know M NaOH and mols NaOH, solve for L NaOH and convert to mL if you wish.
Can you please plug in the numbers this is so confusing.
To find the molarity of the NaOH solution, we need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
(a) To find the moles of NaOH, we need to convert the mass of NaOH to moles using the molar mass.
Molar mass of NaOH = (23.0 g/mol Na) + (16.0 g/mol O) + (1.0 g/mol H) = 40.0 g/mol
moles of NaOH = mass of NaOH / molar mass of NaOH
= 4.315 g / 40.0 g/mol
= 0.1079 mol
Next, we need to convert the volume of the solution to liters. The graduated flask has a volume of 100 mL, which is equivalent to 0.100 L.
Now, we can calculate the molarity:
Molarity = moles of NaOH / volume of solution
= 0.1079 mol / 0.100 L
= 1.08 M
Therefore, the molarity of the NaOH solution is 1.08 M.
(b) To find the volume of the NaOH solution required to neutralize the HCl solution, we need to use the balanced equation to determine the mole ratio between NaOH and HCl.
According to the balanced equation, it tells us that 1 mole of NaOH reacts with 1 mole of HCl. Therefore, the moles of NaOH in the reaction will be equal to the moles of HCl.
moles of HCl = molarity of HCl x volume of HCl (in liters)
= 0.1500 M x 0.02500 L
= 0.00375 mol HCl
Since 1 mole of NaOH reacts with 1 mole of HCl, we need 0.00375 mol of NaOH for the reaction.
Now we can use the molarity to calculate the volume of the NaOH solution:
Volume of NaOH solution = moles of NaOH / molarity
= 0.00375 mol / 1.08 M
= 0.00347 L or 3.47 mL
Therefore, approximately 3.47 mL of the NaOH solution is required to neutralize exactly 25.00 mL of 0.1500 M HCl solution.