# What is the concentration of the predominant form of "H"_3"A" at pH 10.5?

## $\text{p} {K}_{1} = 9.14$ $\text{p} {K}_{2} = 11.7$ $\text{p} {K}_{1} = 13.8$

Jan 29, 2015

The concentration of the dominant form at pH = 10.5 is $\text{6.91 mmol/L}$.

So, you are dealing with a triprotic acid. Here are the three generic forms of the acid

${H}_{3} A + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{2} {A}^{-} + {H}_{3}^{+} O$,  -> "pKa"_1 = 9.14
${H}_{2} {A}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H {A}^{2 -} + {H}_{3}^{+} O$, $\to \text{pKa} = 12.7$
$H {A}^{2 -} + {H}_{2} O r i g h t \le f t h a r p \infty n s {A}^{3 -} + {H}_{3}^{+} O$. $\to \text{pKa} = 13.8$

The pH of the solution is $\text{10.5}$, which means that the dominant species form of the acid will be $H {A}^{2 -}$. This happens because ${\text{pKa"_1 < "pH" < "pKa}}_{2}$.

Now you have to set up a system of three equations with three unknowns - the concentrations of the acid forms. From now on, I'll use this notation for simplicity

$\left[{H}_{3} A\right] = a$, $\left[{H}_{2} {A}^{-}\right] = b$, and $\left[H {A}^{2 -}\right] = c$

$a + b + c = 7.25$ (1)
$p H = p K {a}_{1} + \log \left(\frac{b}{a}\right)$ (2)
$p H = p K {a}_{2} + \log \left(\frac{c}{b}\right)$ (3)

Plugging in the pH and the two pKa values will get

$10.5 = 9.14 + \log \left(\frac{c}{b}\right)$
$10.5 = 12.7 + \log \left(\frac{b}{a}\right)$

Solving these two equations for $\frac{c}{b}$ and $\frac{b}{a}$ will get you

$\frac{c}{b} = 0.0063$ and $\frac{b}{a} = 22.9$

SInce we need to determine the value of $b$, we'll plug

$c = 0.0063 \cdot b$ and $a = \frac{b}{22.9}$ into equation (1), which will give

$\frac{b}{22.9} + b + 0.0063 b = 7.25$

Finally, you'll get $b = 6.91$.

Therefore, the concentration of ${H}_{2} {A}^{-}$, the dominant form of the acid, will be

$\left[{H}_{2} {A}^{-}\right] = \text{6.91 mmol/L}$

Jan 30, 2015

The dominant form at pH 10.5 is H₂A⁻, with a concentration of 6.9 mmol/L.

Warning: This is a very long answer, but this is a complicated problem.

We must use the systematic approach to chemical equilibrium.

H₃A ⇌ H⁺ + H₂A⁻; ${K}_{1}$
H₂A⁻ ⇌ H⁺ + HA²⁻; ${K}_{2}$
HA²⁻ ⇌ H⁺ + A³⁻; ${K}_{3}$
H₂O = H⁺ + OH⁻; ${K}_{\text{w}}$

Species:

H₃A, H₂A⁻, HA²⁻, A³⁻, H⁺, OH⁻ (6 species)

I am going to omit the charges and concentration brackets for easier typing. Only 2 significant figures are justified, but I will carry 3 to avoid round-off errors. I will round off at the end.

We can't use charge balance, because the pH is fixed.

Mass Balance:

(1) $0.007 25 = \text{H"_3"A" + "H"_2"A" + "HA + A}$
(2) "H" = 10^-10.5 = 3.16 × 10^-11

(3) K_1 = ("H × H"_2"A")/("H"_3"A") = 10^-9.14 = 7.24 × 10^-10
(4) K_2 = "H × HA"/("H"_2"A") = 10^-12.7 = 2.00 × 10^-13
(5) K_3 = "H × A"/"HA" = 10^-13.8 = 1.58 × 10^-14
(6) K_"w" = "H × OH" = 1.00 × 10^-14

There are six equations and six unknowns. We're in business.

SOLUTION:

From (2),
(7) "H" = 3.16 × 10^-11

From (6), "OH" = K_"w"/"H" = (1.00 × 10^-14)/( 3.16 × 10^-11)

(8) "OH" = 3.16 × 10^-4

Since the pH is between $\text{p} {K}_{1}$ and $\text{p} {K}_{2}$, we will assume that the third step is negligible and that $A$ ≈ 0.

Then, from (1),

(9) $0.007 25 = \text{H"_3"A" + "H"_2"A" + "HA}$

From (3), K_1 × "H"_3"A" = "H × H"_2"A"

(10) $\text{H"_2"A" = (K_1 × "H"_3"A") /"H}$

From (4), K_2 × "H"_2"A" = "H × HA"

(11) "H"_2"A" = (K_1 × "HA") /K_2

From (10) and (11), (K_1 × "H"_3"A")/"H" = "H× HA"/K_2

(12) ${\text{HA" = (K_1K_2"H"_3"A")/"H}}^{2}$

Substitute (10) and (12) in (9).

$0.007 25 = {\text{H"_3"A" + (K_1"H"_3"A")/"H" + (K_1K_2"H"_3"A")/"H}}^{2}$

0.007 25 = "H"_3"A"(1 + K_1/"H" + (K_1K_2)/"H"^2)

0.007 25 = "H"_3"A"(1 + (7.24 × 10^-10)/(3.16 × 10^-11) + (7.24 × 10^-10 × 2.00 × 10^-13)/((3.16 × 10^-11)^-2))

$0.007 25 = \text{H"_3"A"(1 + 22.9 + 0.145) = 24.0"H"_3"A}$

$\text{H"_3"A} = \frac{0.00725}{24.0}$

(13) "H"_3"A" = 3.02 × 10^-4

From (10), "H"_2"A" = (K_1 × "H"_3"A")/("H") = (7.24 × 10^-10 × 3.02 × 10^-4)/(3.16 × 10^-11)

(14) "H"_2"A" = 6.92 × 10^-3

From (12), "HA" = (K_1K_2"H"_3"A")/"H"^2 = (7.24 × 10^-10 × 2.00 × 10^-13 × 3.02 × 10^-4)/(3.16 × 10^-11)^2

(15) "HA" = 4.38 × 10^-5

From (5), "A" = (K_3"HA")/"H" = (1.58 ×10^-14 × 4.38 × 10^-5)/(3.16 × 10^-11)

(16) "A" = 2.19 × 10^-8

In decreasing order of concentration,

[H₂A⁻] = 6.9 × 10⁻³ mol/L = 6.9 mmol/L
[OH⁻] = 3.2 × 10⁻⁴ mol/L
[H₃A] = 3.0 × 10⁻⁴ mol/L
[HA²⁻] = 4.4 × 10⁻⁵ mol/L
[A³⁻] = 2.2 × 10⁻⁸ mol/L
[H⁺] = 3.2 × 10⁻¹¹ mol/L

Check: [H₃A] + [H₂A⁻] + [HA²⁻] + [A³⁻] = (3.0 × 10⁻⁴ + 6.9 × 10⁻³ + 4.4 × 10⁻⁵ + 2.2 × 10⁻⁸) mol/L = 7.2 × 10⁻³ mol/L = 7.2 mmol/L