24) It was found that a 20.00 mL portion of a solution of oxalic acid, H2C2O4, requires 6.69 mL of 0.200 M K2Cr2O7 for complete reaction in an acidic solution. In the reaction, the oxidation product is CO2 and the reduction product is Cr3+. How many milliliters of 0.450 M NaOH are required to completely neutralize the H2C2O4 in a separate 20.00 mL sample of the same oxalic acid solution?
Balance the redox part first. You don't need to write the entire equation, just the redox part.
C2O4^2- + Cr2O7^2- ==> 2Cr^3+ + 2CO2
C in C2O4 changes from +6 (for 2C) to +8 (for 2C) which is a change of 2e.
Cr in Cr2O7^2- changes from +12 (for 2Cr) to +6(for 2 Cr) which is a change of 6e so the balanced redox part is
3C2O4^2- + Cr2O7^2- ==> 2Cr^3+ + 6CO2
millimols Cr2O7^2- = mL x M = estimated 1.3.
mmols C2O4^2- = estimated 1.3* 3 = estimated 3.9
M C2O4^2- = mmols/mL = estimated 3.9/20 = estd 0.2M
Then the acid/base titration is
H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O
mmols H2C2O4 = mL x M = 20 x 0.2 = estd 4
mmols NaOH = 2*4 = estd 8
M = mmols/mL or mL = mmols/M = estd 8/0.450 = ? mL NaOH
Thank you!
To determine the volume of 0.450 M NaOH needed to neutralize the H2C2O4 in a separate 20.00 mL sample of the same oxalic acid solution, we can use the stoichiometry of the reaction.
The balanced equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is as follows:
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
From the equation, we can see that the stoichiometric ratio between H2C2O4 and NaOH is 1:2. This means that for every 1 mole of H2C2O4, we need 2 moles of NaOH.
First, let's determine the number of moles of H2C2O4 in the 20.00 mL sample of the oxalic acid solution.
To do this, we'll use the concentration of the potassium dichromate (K2Cr2O7) solution (0.200 M) and the volume required (6.69 mL) to react with the oxalic acid.
The balanced equation for the reaction between K2Cr2O7 and H2C2O4 is:
H2C2O4 + 3 K2Cr2O7 + 8 H2SO4 → 3 CO2 + 3 Cr2(SO4)3 + 4 K2SO4 + 8 H2O
From the equation, we can see that the stoichiometric ratio between K2Cr2O7 and H2C2O4 is 3:1. This means that for every 3 moles of K2Cr2O7, we need 1 mole of H2C2O4.
We'll use the equation:
(0.200 mol/L) × (6.69 mL) = (x mol/L) × (20.00 mL)
where x is the number of moles of H2C2O4.
Rearranging the equation, we find:
x = (0.200 mol/L) × (6.69 mL) / (20.00 mL)
x = 0.0669 mol
So, we have 0.0669 mol of H2C2O4 in the 20.00 mL sample.
Since the stoichiometric ratio between H2C2O4 and NaOH is 1:2, we'll need twice the number of moles of NaOH to neutralize the H2C2O4.
The number of moles of NaOH needed is:
2 * (0.0669 mol) = 0.1338 mol
To determine the volume of 0.450 M NaOH required, we'll use the equation:
(0.450 mol/L) × (V mL) = (0.1338 mol) × (20.00 mL)
Solving for V, we find:
V = (0.1338 mol) × (20.00 mL) / (0.450 mol/L)
V ≈ 5.956 mL
Therefore, approximately 5.956 mL of 0.450 M NaOH is required to completely neutralize the H2C2O4 in the separate 20.00 mL sample of the oxalic acid solution.
To calculate the amount of NaOH needed to neutralize the oxalic acid, we need to consider the stoichiometry of the reaction between oxalic acid (H2C2O4) and NaOH. This reaction can be represented by the following balanced equation:
2H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
From the equation, we can see that 2 moles of oxalic acid react with 2 moles of NaOH to form 1 mole of sodium oxalate and 2 moles of water.
Given that the volume of the oxalic acid solution is 20.00 mL and the concentration of NaOH is 0.450 M, we can use the following equation to calculate the amount of NaOH needed:
moles of H2C2O4 = volume (in L) × concentration (in mol/L)
moles of NaOH = (moles of H2C2O4 / 2)
volume of NaOH = (moles of NaOH / concentration of NaOH)
Now let's calculate the moles of H2C2O4:
moles of H2C2O4 = 20.00 mL × (1 L / 1000 mL) × (6.69 mL K2Cr2O7 / 20.00 mL H2C2O4) × (0.200 mol K2Cr2O7 / 1 L)
= 0.002014 mol H2C2O4
Next, we calculate the moles of NaOH:
moles of NaOH = (0.002014 mol H2C2O4 / 2)
= 0.001007 mol NaOH
Finally, we calculate the volume of NaOH needed:
volume of NaOH = (0.001007 mol NaOH / 0.450 mol/L)
= 0.002238 L or 2.24 mL (rounded to two decimal places)
Therefore, approximately 2.24 mL of 0.450 M NaOH are required to completely neutralize the H2C2O4 in a separate 20.00 mL sample of the same oxalic acid solution.