# Question #4f094

Apr 26, 2015

If you want the concentration of all the species present in solution, you'll need the acid dissociation constants for phosphoric acid's second and third ionizations.

These are listed as being

${K}_{a 2} = 6.2 \cdot {10}^{- 8}$ and ${K}_{a 3} = 4.2 \cdot {10}^{- 13}$.

As you can see, these acid dissociation constants have very, very small values, which is why only the first ionization of phophoric acid is usually taken into consideration in pH calculations.

So, the first equilibrium reaction that is established in solution will be

$\text{ } {H}_{3} P {O}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + {H}_{2} P {O}_{4 \left(a q\right)}^{-}$,
I.......0.02.........................................0......................0
C......(-x)..........................................(+x)..................(+x)
E...0.02-x.........................................x........................x

By definition, the acid dissociation constant is

${K}_{a 1} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[{H}_{2} P {O}_{4}^{-}\right]}{\left[{H}_{3} P {O}_{4}\right]} = \frac{x \cdot x}{0.02 - x} = {x}^{2} / \left(0.02 - x\right)$

Solving for $x$ will produce a positive and a negative solution. Since $x$ represents concentration, the negative solution is automatically eliminated., which leaves x = 0.009060.

As a result, $\left[{H}_{3} {O}^{+}\right] = \left[{H}_{2} P {O}_{4}^{-}\right] = \text{0.009060 M}$

Now for the second equilibrium reaction

$\text{ } {H}_{2} P {O}_{4 \left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + H P {O}_{4 \left(a q\right)}^{2 -}$
I...0.009060..............................0.00906............0
C......(-x)..........................................(+x)...............(+x)
E...0.009060-x.......................0.009060+x.......x

${K}_{a 2} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[H P {O}_{4}^{2 -}\right]}{\left[H 2 P {O}_{4}^{-}\right]} = \frac{\left(0.009060 + x\right) \cdot x}{0.009060 - x}$

Because the second acid dissociation constant is so small, you can approximate 0.009060 + x and 0.009060 - x with 0.009060. The above equation becomes

$6.2 \cdot {10}^{- 8} = \frac{\cancel{0.009060} \cdot x}{\cancel{0.009060}} = x$

As a result, $\left[H P {O}_{4}^{2 -}\right] = 6.2 \cdot {10}^{- 8} \text{M}$

$\left[{H}_{3} {O}^{+}\right] = 0.009060 + 6.2 \cdot {10}^{- 8} \cong \text{0.009060 M}$

Finally, the third equilibrium reaction

$\text{ } H P {O}_{4 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + P {O}_{4 \left(a q\right)}^{3 -}$
I...$6.2 \cdot {10}^{- 8}$...........................0.009060.........0
C.......(-x).............................................(+x)..........(+x)
E...$6.2 \cdot {10}^{- 8} - x$.................0.009060+x.......x

${K}_{a 3} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[P {O}_{4}^{30}\right]}{\left[H P {O}_{4}^{2 -}\right]} = \frac{\left(0.009060 + x\right) \cdot x}{6.2 \cdot {10}^{- 8} - x}$

Once again, you can safely approximate the above ratio with

$4.2 \cdot {10}^{- 13} = \frac{0.009060 \cdot x}{6.2 \cdot {10}^{- 8}} \implies x = 2.8 \cdot {10}^{- 18}$

As a result, $\left[P {O}_{4}^{3 -}\right] = 2.9 \cdot {10}^{- 18} \text{M}$

$\left[{H}_{3} {O}^{+}\right] = 0.009060 + 2.9 \cdot {10}^{- 18} \cong \text{0.009060 M}$

Therefore, the concentration of all the species listed will be

$\left[{H}_{3} P {O}_{4}\right] = 0.02 - 0.009060 = \textcolor{g r e e n}{\text{0.01094 M}}$
$\left[{H}_{2} P {O}_{4}^{-}\right] = \textcolor{g r e e n}{\text{0.009060 M}}$
$\left[H P {O}_{4}^{2 -}\right] = \textcolor{g r e e n}{6.2 \cdot {10}^{- 8} \text{M}}$
$\left[P {O}_{4}^{3 -}\right] = \textcolor{g r e e n}{2.9 \cdot {10}^{- 18} \text{M}}$
$\left[{H}_{3} {O}^{+}\right] = \textcolor{g r e e n}{\text{0.009060 M}}$