H2C2O4.2H2O(s) is primary standard substance. 2.3688g of oxalic acid hydrate were completely neutralized by 42.56ml of NaOH solution. Calculate the molar concentration of the NaOH solution. Write the balanced equation for the reaction.
H2C2O4.H2O + 2NaOH ==> Na2C2O4 + 3H2O
mols H2C2O4.H2O = grams/molar mass
mols NaOH = twice that (look at the coefficients in the balanced equation).
Then M NaOH = mols NaOH/L NaOH.
Oxalic acid dihydrate:
H2C2O4.2H2O(s) same as H6C2O6
H6C2O6(s) + 2NaOH(aq)--> 4H2O + Na2C2O4(aq)
Moles of NaOH:
(2.3688g H6C2O6)*
(1mol H6C2O6/126.064g H6C2O6)*(2mol NaOH/1 mol H6C2O6)
=.037581mol NaOH
Liters of NaOH:
42.56ml= .004256L
NaOH Molarity(molar concentration)
.037581mol NaOH/.004256L NaOH
=8.830M
Well, first things first, let's unmask the clown within me and get serious for a moment. We can start by writing the balanced equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH):
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
Now let's tackle the calculations. We know that 2.3688 g of oxalic acid hydrate were completely neutralized by 42.56 mL of NaOH solution. To find the molar concentration of NaOH, we need to determine how many moles of oxalic acid were present.
The molar mass of oxalic acid hydrate (H2C2O4·2H2O) is:
[(2 × 1.00784) + (2 × 12.0107) + (4 × 15.999) + 2 × (2 × 1.00784) + 2 × (1.00784 + 15.999)] g/mol
= 126.064 g/mol
Using the molar mass and the given mass of oxalic acid hydrate (2.3688 g), we can calculate the number of moles:
moles of oxalic acid = 2.3688 g / 126.064 g/mol
Now, we need to determine the number of moles of NaOH used in the reaction. Since the reaction is a 1:2 ratio between H2C2O4 and NaOH, the moles of NaOH will be twice the moles of oxalic acid.
moles of NaOH = 2 × moles of oxalic acid
Finally, we can find the molar concentration (in mol/L or M) of NaOH by dividing the moles of NaOH by the volume of NaOH solution in liters:
molar concentration (NaOH) = moles of NaOH / volume of NaOH solution (in L)
And there you have it, the molar concentration of the NaOH solution!
To calculate the molar concentration of NaOH solution, we need to determine the number of moles of oxalic acid hydrate (H2C2O4.2H2O) and the stoichiometry of the reaction with NaOH.
The balanced equation for the reaction between oxalic acid (H2C2O4) and NaOH is:
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
From the equation, we can see that 1 mole of oxalic acid reacts with 2 moles of NaOH.
Step 1: Calculate the number of moles of oxalic acid hydrate:
Given mass of oxalic acid hydrate = 2.3688 g
Molar mass of H2C2O4 . 2H2O = 126 g/mol
Number of moles of oxalic acid hydrate = mass / molar mass
= 2.3688 g / 126 g/mol
Step 2: Calculate the number of moles of NaOH:
From the balanced equation, we know that 1 mole of oxalic acid reacts with 2 moles of NaOH.
Therefore, the number of moles of NaOH is the same as the number of moles of oxalic acid.
Number of moles of NaOH = 2.3688 g / 126 g/mol
Step 3: Calculate the molar concentration of NaOH solution:
Molar concentration (M) = Number of moles / Volume (L)
Given volume of NaOH solution = 42.56 mL = 42.56 / 1000 = 0.04256 L
Molar concentration of NaOH solution = Number of moles / Volume
= (2.3688 g / 126 g/mol) / 0.04256 L
Calculate the value to determine the molar concentration of the NaOH solution.
Finally, the balanced equation for the reaction between oxalic acid and NaOH is:
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
To calculate the molar concentration of the NaOH solution, we need to use the equation and stoichiometry of the reaction between oxalic acid and NaOH.
The balanced equation for the reaction between oxalic acid (H2C2O4) and NaOH is:
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
From the balanced equation, we can see that one mole of oxalic acid (H2C2O4) reacts with two moles of NaOH to produce one mole of sodium oxalate (Na2C2O4) and two moles of water (H2O).
Given that 2.3688g of oxalic acid hydrate (H2C2O4.2H2O) was completely neutralized, we need to convert this mass to moles. The molar mass of H2C2O4.2H2O can be calculated as follows:
Molar mass (H2C2O4.2H2O) = (atomic mass of H × 2) + (atomic mass of C × 2) + (atomic mass of O × 4) + (molar mass of H2O × 2)
= (1.01 g/mol × 2) + (12.01 g/mol × 2) + (16.00 g/mol × 4) + (18.02 g/mol × 2)
= 126.07 g/mol
Now we can calculate the number of moles of H2C2O4.2H2O:
Number of moles = Mass (g) / Molar mass (g/mol)
= 2.3688 g / 126.07 g/mol
Next, we need to determine the number of moles of NaOH that reacted with the oxalic acid. According to the stoichiometry of the balanced equation, 1 mole of H2C2O4 reacts with 2 moles of NaOH. Therefore, the number of moles of NaOH is:
Number of moles of NaOH = (Number of moles of H2C2O4.2H2O) × (2 moles of NaOH / 1 mole of H2C2O4)
Finally, we can calculate the molar concentration of the NaOH solution:
Molar concentration (NaOH) = Number of moles of NaOH / Volume of NaOH solution (in liters)
The volume of NaOH solution is given as 42.56 mL, so we need to convert it to liters:
Volume (L) = 42.56 mL / 1000 mL/L
Now, plug the values into the equation to calculate the molar concentration of the NaOH solution.