# What is the pH of a solution that is 5.0×10−2 M in H2CO3?

Apr 8, 2018

See below:

#### Explanation:

${H}_{2} C {O}_{3}$, or carbonic acid, is a weak acid formed from carbon dioxide reacting with water.

$C {O}_{2} \left(g\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}_{2} C {O}_{3} \left(a q\right)$

Being a weak acid, it will only partially dissociate in water, and has a dissociation constant, ${K}_{a}$, of $4.3 \times {10}^{-} 7$ according to This table. Really, carbonic acid is diprotic, meaning it can dissociate twice, so we have a second ${K}_{a}$ value for the second dissociation: ${K}_{a} = 4.8 \times {10}^{-} 11$. Which will also contribute to the $p H$. (albeit to a smaller extent than the first dissociation)

Lets set up the dissociation equation for ${K}_{a}$ of the first dissociation:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[H C {O}_{3}^{-}\right]}{\left[{H}_{2} C {O}_{3}\right]}$

Now, let's plug in our values for the concentration of the carbonic acid, along with the ${K}_{a}$ value.

$4.3 \times {10}^{-} 7 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[H C {O}_{3}^{-}\right]}{5.0 \times {10}^{-} 2}$

$2.15 \times {10}^{-} 8 = \left(\left[{H}_{3} {O}^{+}\right] \times \left[H C {O}_{3}^{-}\right]\right)$

Now, we can assume that $\left[{H}_{3} {O}^{+}\right] = \left[H C {O}_{3}^{-}\right]$ as they exist in a 1:1 ratio in the solution. This allows to take the square root out of the expression $\left(\left[{H}_{3} {O}^{+}\right] \times \left[H C {O}_{3}^{-}\right]\right)$ to find the respective concentrations:

$\sqrt{2.15 \times {10}^{-} 8} \approx \left(1.47 \times {10}^{-} 4\right) = \left(\left[{H}_{3} {O}^{+}\right] = \left[H C {O}_{3}^{-}\right]\right)$

Now, in the second dissociation, the $\left[H C {O}_{3}^{-}\right]$ ion will act as the acid, and therefore the concentration of this species, which we found in the first dissociation, will be the value of the denominator in the new ${K}_{a}$ expression:

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[C {O}_{3}^{2 -}\right]}{\left[H C {O}_{3}^{-}\right]}$

$4.8 \times {10}^{-} 11 = \frac{\left[{H}_{3} {O}^{+}\right] \times \left[C {O}_{3}^{2 -}\right]}{\left[1.47 \times {10}^{-} 4\right]}$

$\approx 7.04 \times {10}^{-} 15 = \left[{H}_{3} {O}^{+}\right] \times \left[C {O}_{3}^{2 -}\right]$

$\sqrt{7.04 \times {10}^{-} 15} \approx 8.39 \times {10}^{-} 8 = \left[{H}_{3} {O}^{+}\right] = \left[C {O}_{3}^{2 -}\right]$

So the concentration of Oxonium ions, $\left[{H}_{3} {O}^{+}\right]$, which determine the $p H$, is approximately $\left(1.47 \times {10}^{-} 4\right) + \left(8.39 \times {10}^{-} 8\right) \approx 1.47 \times {10}^{-} 4$.
In other words, the second dissociation was so small that it could be have been considered negligible- But let's be thorough.

Now, using the equation to find $p H$ we can calculate the $p H$ of this solution.

$p H = - \log \left[{H}_{3} {O}^{+}\right]$
$p H = - \log \left[1.47 \times {10}^{-} 4\right]$
$p H \approx 3.83$