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

The concentration of the dominant form at pH = 10.5 is #"6.91 mmol/L"#.

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

#H_3A + H_2O rightleftharpoons H_2A^(-) + H_3^(+)O#, # -> "pKa"_1 = 9.14#
#H_2A^(-) + H_2O rightleftharpoons HA^(2-) + H_3^(+)O#, # -> "pKa" = 12.7#
#HA^(2-) + H_2O rightleftharpoons A^(3-) + H_3^(+)O#. #-> "pKa" = 13.8#

The pH of the solution is #"10.5"#, which means that the dominant species form of the acid will be #HA^(2-)#. This happens because #"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

#[H_3A] = a#, #[H_2A^(-)] = b#, and #[HA^(2-)] = c#

#a + b + c = 7.25# (1)
#pH = pKa_1 + log(b/a)# (2)
#pH = pKa_2 + log(c/b)# (3)

Plugging in the pH and the two pKa values will get

#10.5 = 9.14 + log(c/b)#
#10.5 = 12.7 + log(b/a)#

Solving these two equations for #c/b# and #b/a# will get you

#c/b = 0.0063# and #b/a = 22.9#

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

#c = 0.0063*b# and #a = b/22.9# into equation (1), which will give

#b/22.9 + b + 0.0063b = 7.25#

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

Therefore, the concentration of #H_2A^(-)#, the dominant form of the acid, will be

#[H_2A^(-)] = "6.91 mmol/L"#

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.

Chemical Equations:

H₃A ⇌ H⁺ + H₂A⁻; #K_1#
H₂A⁻ ⇌ H⁺ + HA²⁻; #K_2#
HA²⁻ ⇌ H⁺ + A³⁻; #K_3#
H₂O = H⁺ + OH⁻; #K_"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 = "H"_3"A" + "H"_2"A" + "HA + A"#
(2) #"H" = 10^-10.5 = 3.16 × 10^-11#

Equilibrium Constants:

(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 #"p"K_1# and #"p"K_2#, we will assume that the third step is negligible and that #A# ≈ 0.

Then, from (1),

(9) #0.007 25 = "H"_3"A" + "H"_2"A" + "HA"#

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

(10) #"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) #"HA" = (K_1K_2"H"_3"A")/"H"^2#

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

#0.007 25 = "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 = "H"_3"A"(1 + 22.9 + 0.145) = 24.0"H"_3"A"#

#"H"_3"A" = 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