Calculate the pH and the poH 0.0035moldm-3 H3 PO4 solution
The first ionization is
.................H3PO4 ==> H^+ + H2PO4^-
I..............0.0035...........0...........0
C...............-x.................x............x
E............0.0035-x..........x............x
k1 = (H^+)(H2PO4^-)/(H3PO4)
Look up k1and plug the E line into the equation. Solve for (H^+) a,d convert to pH and pOH.
NOTE 1: You can ignore the H^+ from k2 and k3.
NOTE 2: Probably you will need to solve the quadratic equation; i.e. you will not be able to ignore x in the 0.0035-x above.
Post your work if you get stuck.
To calculate the pH and pOH of a solution of H3PO4, we need to determine the concentration of H+ ions in the solution.
H3PO4 is a triprotic acid, which means it can donate three hydrogen ions (H+). Each dissociation step has its own dissociation constant (Ka).
The dissociation reactions of H3PO4 are as follows:
H3PO4 ⇌ H+ + H2PO4-
H2PO4- ⇌ H+ + HPO42-
HPO42- ⇌ H+ + PO43-
Let's calculate the concentration of H+ ions for each dissociation step:
1. For the first dissociation step:
[H+] = [H2PO4-] = Ka1 × [H3PO4]
Where Ka1 is the first dissociation constant of H3PO4. The value of Ka1 for H3PO4 is approximately 7.5 x 10^-3 mol/dm^3.
[H+] = 7.5 x 10^-3 mol/dm^3 × 0.0035 mol/dm^3
[H+] = 2.625 x 10^-5 mol/dm^3
2. For the second dissociation step:
[H+] = [HPO42-] = Ka2 × [H2PO4-]
Where Ka2 is the second dissociation constant of H3PO4. The value of Ka2 for H3PO4 is approximately 6.2 x 10^-8 mol/dm^3.
[H+] = 6.2 x 10^-8 mol/dm^3 × 2.625 x 10^-5 mol/dm^3
[H+] = 1.6275 x 10^-12 mol/dm^3
3. For the third dissociation step:
[H+] = [PO43-] = Ka3 × [HPO42-]
Where Ka3 is the third dissociation constant of H3PO4. The value of Ka3 for H3PO4 is approximately 4.8 x 10^-13 mol/dm^3.
[H+] = 4.8 x 10^-13 mol/dm^3 × 1.6275 x 10^-12 mol/dm^3
[H+] = 7.801 x 10^-25 mol/dm^3
Since the pH is defined as the negative logarithm (base 10) of the concentration of H+ ions, we can calculate the pH using the formula:
pH = -log10[H+]
pH = -log10(7.801 x 10^-25)
pH ≈ 24.11
The pOH can be calculated using the formula:
pOH = 14 - pH
pOH = 14 - 24.11
pOH ≈ -10.11
Therefore, the pH of the 0.0035 mol/dm3 H3PO4 solution is approximately 24.11 and the pOH is approximately -10.11.
To calculate the pH and pOH of a solution of H3PO4, you need to know the concentration of the solution and the dissociation constants of the acid.
H3PO4 is a polyprotic acid, which means it can lose multiple protons. In this case, it can lose one, two, or three protons, depending on the conditions. The dissociation reactions are as follows:
H3PO4 ⇌ H+ + H2PO4-
H2PO4- ⇌ H+ + HPO42-
HPO42- ⇌ H+ + PO43-
The concentration given is 0.0035 moldm-3. Since all three dissociation reactions occur, we need to consider them all. However, because H3PO4 is a weak acid and does not dissociate completely, we can approximate that the concentration of H+ ions (or H3O+ ions, which are essentially the same) is equal to the concentration of the acid (0.0035 moldm-3) after the first dissociation reaction.
So, the [H+] concentration in the solution is 0.0035 moldm-3.
To calculate the pH and pOH, we can use the following formulas:
pH = -log[H+]
pOH = -log[OH-]
But in this case, since we are dealing with an acid, and not a base, we don't really need the pOH value.
Using the concentration obtained earlier, we can calculate the pH as follows:
pH = -log(0.0035) = 2.46
Therefore, the pH of the 0.0035 moldm-3 H3PO4 solution is approximately 2.46.