an aqeous sollution of 6.3 g oxalic acid is made up to 250 mL. calculate the volume of 0.1 N sodium hydroxide needed to neutralise 10 mL of this sollution?
As per previous question, the stoichiometry gives:
m/w = cV/2
However, this time your given the mass of oxalic acid, in the aloquat, and wish to find V.
Given:
m = (6.3)/25 g
c = 0.1N = 0.1M
w = 90.03517 g/mol
V = 2m/cw
40 mL
To calculate the volume of 0.1 N sodium hydroxide needed to neutralize 10 mL of the given solution, we need to first determine the molarity of the oxalic acid solution.
The molarity (M) of a solution is calculated using the formula:
Molarity (M) = (moles of solute) / (volume of solution in liters)
Given that 6.3 g of oxalic acid is dissolved in 250 mL of solution, we need to convert the grams of oxalic acid to moles. The molar mass of oxalic acid (H2C2O4) is:
2 * (1.01 g/mol) + 2 * (12.01 g/mol) + 4 * (16.00 g/mol) = 90.03 g/mol
Using this molar mass, we calculate the number of moles:
moles of oxalic acid = (mass of oxalic acid) / (molar mass of oxalic acid)
= 6.3 g / 90.03 g/mol
Now, we need to convert the volume of the solution from milliliters to liters:
volume of solution = 250 mL = 250 mL * (1 L / 1000 mL)
Next, we use the molarity formula to determine the molarity of the oxalic acid solution:
Molarity (M) = (moles of solute) / (volume of solution in liters)
= (moles of oxalic acid) / (volume of solution)
Now that we have the molarity of the oxalic acid solution, we can calculate the volume of 0.1 N sodium hydroxide needed to neutralize 10 mL of the solution.
The term "0.1 N" represents the molarity of the sodium hydroxide. Since the question asks for the volume required to neutralize 10 mL of the solution, we can set up the following equation:
(Molarity of oxalic acid solution) * (volume of oxalic acid solution) = (Molarity of sodium hydroxide) * (volume of sodium hydroxide)
substituting the values:
unknown * 10 mL = 0.1 N * unknown
simplifying the equation:
unknown * 0.01 L = 0.1 N * unknown
Since the unknown volume value appears on both sides of the equation, we can eliminate it:
0.01 L = 0.1 N
Therefore, the volume of 0.1 N sodium hydroxide needed to neutralize 10 mL of the given solution is 0.01 L or 10 mL.