Calculate the molarity of a sodium hydroxide solution if 10.42 mL of this solution are needed to neutralize 25.00 mL of 0.2043 M oxalic acid. how do you do the working out

Well, aren't we in for a fun chemistry calculation! Let's clown around a bit and solve this problem step by step.

First, let's start with what we know. We have 10.42 mL of sodium hydroxide solution (NaOH) that neutralizes 25.00 mL of 0.2043 M oxalic acid (H2C2O4).

To find the molarity of the NaOH solution, we need to use the stoichiometry of the reaction. The balanced equation for the reaction is:

2 NaOH + H2C2O4 -> Na2C2O4 + 2 H2O

Since the ratio of NaOH to H2C2O4 is 2:1, we know that for every mole of oxalic acid, we need 2 moles of sodium hydroxide.

Now, let's do some math! We know that the molarity (M) is defined as moles of solute per liter of solution.

Step 1: Calculate the moles of oxalic acid

Moles of oxalic acid = Volume (in L) * Molarity
= 0.02500 L * 0.2043 mol/L
= 0.0051075 moles of oxalic acid

Step 2: Calculate the moles of sodium hydroxide

Since we have a 2:1 mole ratio, the moles of sodium hydroxide will be twice the moles of oxalic acid.

Moles of NaOH = 2 * Moles of oxalic acid
= 2 * 0.0051075 moles
= 0.010215 moles of NaOH

Step 3: Calculate the molarity of the NaOH solution

Molarity = Moles of NaOH / Volume (in L)
= 0.010215 moles / 0.01042 L
= 0.9803 M (approximately)

So, the molarity of the sodium hydroxide (NaOH) solution is approximately 0.9803 M.

Hope this explanation brought a smile to your face! If you have any more questions, feel free to ask this jester of chemistry.

2NaOH + H2C2O4 ==> Na2C2O4 + 2H2O

mols H2C2O4 = M x L = ?
mols NaOH = 1/2 that (look at the coefficients in the balanced equation.)
M NaOH = mols NaOH/L NaOH.

0.9808 M NaOH

To calculate the molarity of the sodium hydroxide (NaOH) solution, we can use the following equation:

M₁V₁ = M₂V₂

Where:
M₁ = molarity of the oxalic acid (given as 0.2043 M)
V₁ = volume of the oxalic acid solution (given as 25.00 mL)
M₂ = molarity of the sodium hydroxide solution (what we need to calculate)
V₂ = volume of the sodium hydroxide solution (given as 10.42 mL)

Now, substitute the given values into the equation and solve for M₂:

(0.2043 M) * (25.00 mL) = M₂ * (10.42 mL)

Multiply the numbers on each side of the equation:

5.1075 = M₂ * 10.42

Next, solve for M₂ by dividing both sides of the equation by 10.42:

M₂ = 5.1075 / 10.42

M₂ ≈ 0.4906 M

Therefore, the molarity of the sodium hydroxide solution is approximately 0.4906 M.

To calculate the molarity (M) of a solution, you need to know the volume (in liters) and the number of moles of the solute. In this case, the solute is sodium hydroxide (NaOH).

Step 1: Find the number of moles of oxalic acid (H2C2O4):
Molarity (M) = moles of solute / volume of solution (in liters)

Given data:
Volume of oxalic acid solution (V1) = 25.00 mL = 0.02500 L
Molarity of oxalic acid solution (M1) = 0.2043 M

moles of H2C2O4 = M1 × V1 = 0.2043 M × 0.02500 L = 0.005108 mol

Step 2: Determine the number of moles of sodium hydroxide (NaOH):
From the balanced chemical equation between NaOH and oxalic acid (H2C2O4), the stoichiometric ratio is 2:1. This means that for every 2 moles of NaOH, you will need 1 mole of H2C2O4.

moles of NaOH = 0.005108 mol × 2 = 0.01022 mol

Step 3: Calculate the molarity of the sodium hydroxide solution:
Now, you know the volume of the sodium hydroxide solution (V2) is 10.42 mL = 0.01042 L. Let's assume the molarity of NaOH is M2.

Molarity (M2) = moles of NaOH / volume of NaOH solution

M2 = 0.01022 mol / 0.01042 L = 0.9822 M

Therefore, the molarity of the sodium hydroxide solution is approximately 0.9822 M.