A phosphate buffer solution contain 0.08 M K2HPO4 and 0.12 M KH2PO4. pKa values for H3PO4 are 2, 6.8, and 12.2. Calculate the concentrations of

a) H3PO4
b) H2PO4-
c) HPO42-
d) H+
e) OH-
f) K+

To calculate the concentrations of the different species in the phosphate buffer solution, we will use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation relates the pH of a solution to the ratio of the concentrations of the acidic and basic forms of a buffer. It is given by:

pH = pKa + log([A-]/[HA])

Where:
- pH is the logarithm of the hydrogen ion concentration ([H+])
- pKa is the logarithm of the acid dissociation constant
- [A-] is the concentration of the conjugate base
- [HA] is the concentration of the acid

a) H3PO4 and H2PO4-:
The pKa values given for H3PO4 are 2, 6.8, and 12.2.

To calculate the concentration of H3PO4 and H2PO4-, we need to consider the dissociation of H3PO4 as three separate reactions:
H3PO4 + H2O ↔ H2PO4- + H3O+
H2PO4- + H2O ↔ HPO42- + H3O+
HPO42- + H2O ↔ PO43- + H3O+

Since the buffer solution contains K2HPO4 and KH2PO4 in a 1:1 ratio, the initial concentrations of H3PO4 and H2PO4- will be the same:

H3PO4: 0.08 M
H2PO4-: 0.12 M

b) H2PO4- and HPO42-:
Using the Henderson-Hasselbalch equation, we can calculate the concentrations of H2PO4- and HPO42- at a given pH.

pH = pKa + log([A-]/[HA])

At pH 2:
pKa = 2
[H2PO4-]/[H3PO4] = 10^(pH - pKa)

[H2PO4-]/0.08 = 10^(2 - 2)
[H2PO4-] = 0.08 M

At pH 7:
pKa = 6.8
[HPO42-]/[H2PO4-] = 10^(pH - pKa)

[HPO42-]/0.08 = 10^(7 - 6.8)
[HPO42-] = 0.08 M

At pH 12:
pKa = 12.2
[PO43-]/[HPO42-] = 10^(pH - pKa)

[PO43-]/0.08 = 10^(12 - 12.2)
[PO43-] = 0.0632 M

c) H+ concentration:
The concentration of H+ can be calculated from the equation:

[H+] = [H3O+] = 10^(-pH)

At pH 2:
[H+] = [H3O+] = 10^(-2)
[H+] = 0.01 M

At pH 7:
[H+] = [H3O+] = 10^(-7)
[H+] = 0.0000001 M

At pH 12:
[H+] = [H3O+] = 10^(-12)
[H+] = 0.000000000001 M

d) OH- concentration:
Since water dissociates into H+ and OH- ions, we can calculate the OH- concentration using the equation:

[OH-] = 10^(-pOH)
pOH = 14 - pH

At pH 2:
pOH = 14 - 2
pOH = 12
[OH-] = 10^(-12)

At pH 7:
pOH = 14 - 7
pOH = 7
[OH-] = 10^(-7)

At pH 12:
pOH = 14 - 12
pOH = 2
[OH-] = 10^(-2)

e) K+ concentration:
The concentration of K+ in the buffer solution is equal to the initial concentration of KH2PO4, which is 0.12 M.

To recap:
a) H3PO4: 0.08 M
b) H2PO4-:
- At pH 2: 0.08 M
- At pH 7: 0.08 M
- At pH 12: 0.12 M
c) HPO42-:
- At pH 2: 0.08 M
- At pH 7: 0.08 M
- At pH 12: 0.0632 M
d) H+:
- At pH 2: 0.01 M
- At pH 7: 0.0000001 M
- At pH 12: 0.000000000001 M
e) OH-:
- At pH 2: 10^(-12)
- At pH 7: 10^(-7)
- At pH 12: 10^(-2)
f) K+: 0.12 M

To calculate the concentrations of the different species in the phosphate buffer solution, we need to consider the dissociation of phosphoric acid (H3PO4) and the acid-base equilibria involved. We can use the Henderson-Hasselbalch equation to determine the concentrations of the acid (H3PO4) and its conjugate base (H2PO4- / HPO42-).

a) H3PO4 concentration:
Since we're given the pKa values for H3PO4, we can determine which equilibrium reaction to consider in the Henderson-Hasselbalch equation. In this case, we use the pKa2 value of 6.8 as it is closest to the pH of the buffer solution.

H3PO4 ⇌ H+ + H2PO4-
pKa2 = 6.8

Using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Let's consider [HA] as the concentration of H3PO4 and [A-] as the concentration of H2PO4-. Since they have a 1:1 ratio in the equation, their concentrations are directly related.

pH = 6.8 + log([H2PO4-] / [H3PO4])

Since we don't know the pH of the buffer solution, we cannot directly calculate [H3PO4]. However, we can rearrange the equation to express [H3PO4] in terms of known values:

[H3PO4] = [H2PO4-] * 10^(pH - pKa)

b) H2PO4- concentration:
To determine the concentration of H2PO4-, we need to consider the initial concentrations of K2HPO4 and KH2PO4 in the buffer solution.

The mole ratio of K2HPO4 to KH2PO4 is 2:3. Given that [K2HPO4] = 0.08 M and [KH2PO4] = 0.12 M, we can calculate the total phosphate concentration:

[HPO42-] + [H2PO4-] = [K2HPO4] + [KH2PO4]
[HPO42-] + [H2PO4-] = 0 + 0.12 M (since K2HPO4 does not contribute H2PO4-)

Since [HPO42-] is negligible compared to [H2PO4-], we can assume [H2PO4-] = 0.12 M.

c) HPO42- concentration:
[HPO42-] + [H2PO4-] = 0.12 M
[HPO42-] + 0.12 M = 0.12 M (since [H2PO4-] = 0.12 M)
[HPO42-] = 0 M

d) H+ concentration:
[H+], in this case, is equal to the concentration of H3PO4 since the acid dissociates into H+ and H2PO4-.

[H+] = [H3PO4] (previously calculated in part a)

e) OH- concentration:
Since we don't know the exact pH of the buffer solution, we cannot directly calculate the concentration of OH-. However, we know that at 25°C, the concentration of OH- and H+ in pure water are equal (1 x 10^-7 M). In a buffer solution, the concentration of OH- should be very low.

f) K+ concentration:
The K+ concentration in the buffer solution is equal to the sum of the concentrations of the potassium salts present:

[K+] = [K2HPO4] + [KH2PO4]
[K+] = 0.08 M + 0.12 M
[K+] = 0.20 M

To summarize:
a) [H3PO4] can be calculated using the Henderson-Hasselbalch equation with pKa2 and pH.
b) [H2PO4-] is equal to the concentration of KH2PO4, which in this example is 0.12 M.
c) [HPO42-] is negligible.
d) [H+] is equal to the concentration of H3PO4.
e) [OH-] is very low in a buffer solution and dependent on pH.
f) [K+] is equal to the sum of the concentrations of the potassium salts.

You can get pH from HH equation. From that calculate pOH. I assume you can get K easily. Then for H3PO4 calculate the fraction in that from from H3PO4 = (H^+)^3/D where D is

(H^+)^3 + k1(H^+)^2 + k1k2(H^+) + k1k2k3