The phosphoric acid in a 100.00 mL sample of a cola drink was titrated with 0.1476 M NaOH. The first equivalence point was detected after 13.32 mL of base added, and the second equivalence point after 28.65 mL. Calculate the concentration of H2PO4- in mol/L. (Hint: if only H3PO4 were present, where would the second equivalence point be?)
I have tried using C1V1 = C2V2, using as many different combination of volumes as I can, but I can't get the answer ( I know what the answer is supposed to be and cannot come up with it)... I think I am missing something that's hidden within the hint but I don't know what that could be...any help would be appreciated.
The first eq point is at 13.32 mL and that titrates the first H of the H3PO4. If the sample contained ONLY H3PO4, then the second eq point would be at 13.32 x 2 or 26.64 mL to titrate the second H. So the difference between 28.65 and 26.64 must be due to the H2PO4^-.
(28.65-26.64/1000) x 0.1476 M = moles H2PO4^- and that is the moles in 100 mL sample. That times 10 will be the moles in L.
To calculate the concentration of H2PO4- in mol/L, we need to consider the titration reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH):
H3PO4 + NaOH → NaH2PO4 + H2O
From the given information, we know the volume of NaOH (V2) required to reach the first equivalence point is 13.32 mL. At this point, all the H3PO4 is neutralized to form NaH2PO4:
H3PO4 + NaOH → NaH2PO4 + H2O
We also know the volume of NaOH (V3) required to reach the second equivalence point is 28.65 mL. At this point, all the NaH2PO4 is neutralized to form Na2HPO4:
NaH2PO4 + NaOH → Na2HPO4 + H2O
However, if only H3PO4 were present, the second equivalence point would correspond to the complete neutralization of all three protons. This would require twice the amount of NaOH compared to the first equivalence point.
So, we can calculate the volume of NaOH required to reach the second equivalence point if only H3PO4 were present:
V2_only_H3PO4 = 2 * V2 = 2 * 13.32 mL
To calculate the concentration of H2PO4- in mol/L, we can use the formula:
C3 = (C2 * V2 - C1 * V1) / (V3 - V2_only_H3PO4)
where:
C1 is the concentration of NaOH in mol/L (given as 0.1476 M)
V1 is the volume of NaOH at the first equivalence point (given as 13.32 mL)
C2 is the concentration of H3PO4 in mol/L (to be determined)
V2 is the volume of NaOH at the first equivalence point (given as 13.32 mL)
V2_only_H3PO4 is the volume of NaOH required to reach the second equivalence point with only H3PO4 present (calculated as 2 * V2)
V3 is the volume of NaOH at the second equivalence point (given as 28.65 mL)
Substituting the given values into the formula, we have:
C3 = (0.1476 M * 13.32 mL - C2 * 13.32 mL) / (28.65 mL - 2 * 13.32 mL)
Now, we need the concentration of NaOH to convert the volume (mL) into moles (mol). The concentration of NaOH is given as 0.1476 M, so:
V1 = 13.32 mL * (1 L / 1000 mL) = 0.01332 L
V2 = 28.65 mL * (1 L / 1000 mL) = 0.02865 L
V2_only_H3PO4 = 2 * V2 = 2 * 0.02865 L
Substituting these values into the formula, we have:
C3 = (0.1476 M * 0.01332 L - C2 * 0.01332 L) / (0.02865 L - 2 * 0.02865 L)
Simplifying the equation:
C3 = (0.001957 M - 0.01332 C2) / (-0.02865 L)
Now, to calculate the concentration of H2PO4-, we need to consider the stoichiometry of the reaction. Since H2PO4- is formed when one H+ (proton) is neutralized, the concentration of H2PO4- is equal to the concentration of NaH2PO4 formed at the first equivalence point.
Therefore, C2 = C3
Substituting this back into the equation, we have:
C3 = (0.001957 M - 0.01332 C3) / (-0.02865 L)
0.02865 C3 = 0.001957 M - 0.01332 C3
0.04197 C3 = 0.001957 M
C3 = 0.0466 M
Therefore, the concentration of H2PO4- in the cola drink is approximately 0.0466 mol/L.
To solve this problem, we need to consider the titration curve and the hint provided. Let's break down the problem step by step:
1. Start by identifying the reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH):
H3PO4 + NaOH -> NaH2PO4 + H2O
2. Note that phosphoric acid has three acidic protons (H+), so it can undergo three successive titrations. Each titration will involve a different proton, resulting in different equivalence points.
3. The hint mentions that we should consider where the second equivalence point would be if only H3PO4 were present. This suggests that the first equivalence point corresponds to titrating the first acidic proton, and the second equivalence point corresponds to titrating the second acidic proton.
4. Let's assume that at the second equivalence point, only H2PO4- (sodium dihydrogen phosphate) is present. This is because the second acidic proton has been neutralized.
H3PO4 -> H2PO4- + H+
5. Now, let's calculate the moles of NaOH used at the second equivalence point. We know the concentration of NaOH (0.1476 M) and the volume used (28.65 mL):
Moles of NaOH = concentration × volume
= 0.1476 M × 0.02865 L
6. Since the balanced equation shows a 1:1 stoichiometric ratio between NaOH and H3PO4, the moles of NaOH used at the second equivalence point are equivalent to the moles of H3PO4 present at the first equivalence point (before the second proton is neutralized).
Moles of H3PO4 (first equivalence point) = moles of NaOH (second equivalence point)
7. Similarly, calculate the moles of NaOH used at the first equivalence point. We know the concentration of NaOH (0.1476 M) and the volume used (13.32 mL):
Moles of NaOH = concentration × volume
= 0.1476 M × 0.01332 L
8. Since the balanced equation shows a 1:1 stoichiometric ratio between NaOH and H3PO4, the moles of NaOH used at the first equivalence point are equivalent to the moles of H3PO4 present (before the first proton is neutralized).
Moles of H3PO4 (before the first equivalence point) = moles of NaOH (first equivalence point)
9. Subtract the moles of H3PO4 at the first equivalence point from the moles at the second equivalence point to find the moles of H2PO4- formed:
Moles of H2PO4- = Moles of H3PO4 (first equivalence point) - Moles of H3PO4 (before the first equivalence point)
10. Now, calculate the concentration of H2PO4- (sodium dihydrogen phosphate) in mol/L:
Concentration = Moles of H2PO4- / Volume of solution in L
Note: The volume of solution can be determined by adding the volumes of NaOH used at both the first and second equivalence points to the volume of the cola drink (100.00 mL).
By following these steps, you should be able to calculate the concentration of H2PO4- in mol/L.