A sample of oxalic acid (H2C2O4, with two acidic protons), of volume 37.09 mL was titrated to the stoichiometric point with 22.41 mL of 0.163 M NaOH(aq). What is the molarity of the oxalic acid?
Answer in units of M.
show me your thinking on this.
If one assumes that the 22.41mL of 0.163M NaOH(aq) was used to titrate through both equivalence points, then 2 moles of NaOH(aq) are used for each mole of H₂C₂O₄(aq) neutralized. That is…
H₂C₂O₄(aq) + 2NaOH(aq) => Na₂C₂O₄(aq) + H₂O(l)
2 moles NaOH > 1 mole H₂C₂O₄
=> 1 moles NaOH = 2 mole H₂C₂O₄ for problem solution
Since Molarity x Volume in Liters of Solution = moles, then …
[Molarity x Volume(L)] NaOH used = 2[Molarity x Volume(L)] H₂C₂O₄ neutralized
(0.163 mol/L NaOH)(0.02241L)] used = 2·M(0.03709L)
M(H₂C₂O₄) = [(0.163)(0.02241)/(2 x 0.03709)]M H₂C₂O₄(aq) = 0.0492M H₂C₂O₄(aq)
To find the molarity of the oxalic acid, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH):
H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
From the balanced equation, we can see that 1 mole of oxalic acid (H2C2O4) reacts with 2 moles of sodium hydroxide (NaOH).
Given:
Volume of oxalic acid (H2C2O4): 37.09 mL
Volume of NaOH: 22.41 mL
Molarity of NaOH: 0.163 M
Step 1: Convert the volumes to liters.
Volume of oxalic acid (H2C2O4) = 37.09 mL = 37.09/1000 = 0.03709 L
Volume of NaOH = 22.41 mL = 22.41/1000 = 0.02241 L
Step 2: Determine the number of moles of NaOH used.
Using the equation:
moles = concentration (M) * volume (L)
Moles of NaOH = 0.163 M * 0.02241 L = 0.00364983 moles
Step 3: Use stoichiometry to find the number of moles of oxalic acid (H2C2O4).
From the balanced equation, we know that 2 moles of NaOH react with 1 mole of H2C2O4.
Moles of H2C2O4 = (0.00364983 moles NaOH) * (1 mole H2C2O4 / 2 moles NaOH) = 0.00182492 moles H2C2O4
Step 4: Calculate the molarity of the oxalic acid.
Molarity = moles of H2C2O4 / volume of H2C2O4 (in liters)
Molarity = 0.00182492 moles / 0.03709 L = 0.04925 M
Therefore, the molarity of the oxalic acid is 0.04925 M.
To find the molarity of oxalic acid, we can use the concept of stoichiometry and the equation of the balanced chemical reaction between oxalic acid and sodium hydroxide.
The balanced chemical equation for the reaction is:
H2C2O4 + 2 NaOH → Na2C2O4 + 2 H2O
From the balanced equation, we can see that the stoichiometric ratio between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is 1:2. This means that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide.
Given that the volume of the oxalic acid sample is 37.09 mL and the volume of the NaOH solution used is 22.41 mL, we need to convert the volumes to moles using their respective concentrations.
Step 1: Convert the volume of NaOH to moles.
Since the concentration of NaOH is given as 0.163 M (moles per liter), we can use the following formula to convert the volume to moles:
moles NaOH = concentration NaOH × volume NaOH (in liters)
moles NaOH = 0.163 M × (22.41 mL / 1000 mL/L) [convert mL to L]
moles NaOH = 0.163 M × 0.02241 L
moles NaOH = 0.00365283 mol
Step 2: Calculate the moles of oxalic acid.
Since the stoichiometric ratio between NaOH and H2C2O4 is 2:1, for every 2 moles of NaOH, 1 mole of H2C2O4 is consumed. Therefore, we can say that the moles of H2C2O4 is half the moles of NaOH.
moles H2C2O4 = 0.00365283 mol / 2
moles H2C2O4 = 0.00182639 mol
Step 3: Calculate the molarity of oxalic acid.
Now we can calculate the molarity of the oxalic acid solution using the equation:
molarity H2C2O4 = moles H2C2O4 / volume H2C2O4 (in liters)
volume H2C2O4 = 37.09 mL / 1000 mL/L [convert mL to L]
volume H2C2O4 = 0.03709 L
molarity H2C2O4 = 0.00182639 mol / 0.03709 L
molarity H2C2O4 = 0.04924 M
Therefore, the molarity of the oxalic acid solution is 0.04924 M.