# Question #dc4c4

Apr 29, 2017

$\frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = 0.0823$

#### Explanation:

Let's start out by writing the balanced equation for the dissociation of carbonic acid:

${H}_{2} C {O}_{3} r i g h t \le f t h a r p \infty n s {H}^{+} + H C {O}_{3}^{-}$

Remember the dissociation constant will be the ratio of the dissociated products (${H}^{+}$ and $H C {O}_{3}^{-}$) to the carbonic acid (${H}_{2} C {O}_{3}$).

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

Looking at this equation, we can divide both sides by $\left[{H}^{+}\right]$ to obtain a ratio between the bicarbonate and the carbonic acid.

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

Since we want the ratio of carbonic acid to bicarbonate, just invert the equation:

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

Now, all we need to do is find $\left[{H}^{+}\right]$ using the definition of pH:

$p H = - \log \left[{H}^{+}\right]$
$\left[{H}^{+}\right] = {10}^{- p H}$
$\left[{H}^{+}\right] = {10}^{-} 7.45$
$\left[{H}^{+}\right] = 3.55 \cdot {10}^{-} 8 M$

Plugging back in, the ratio of carbonic acid to bicarbonate becomes:

$\frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = \frac{\left[{H}^{+}\right]}{K} _ d$
$\frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = \frac{3.55 \cdot {10}^{-} 8}{4.31 \cdot {10}^{-} 7}$
$\frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = 0.0823$

Apr 29, 2017

One can also use the Henderson-Hasselbalch Equation to calculate the Bicarb:Carbonic Acid Ratio then take the reciprocal.

#### Explanation:

$p H = p K a + \log \left(\frac{H C {O}_{3}^{-}}{{H}_{2} C {O}_{3}}\right)$

$7.45 = - \log \left(4.31E-7\right)$ + $\log \left(\frac{H C {O}_{3}^{-}}{{H}_{2} C {O}_{3}}\right)$

$7.45 = 6.37 + \log \left(\frac{H C {O}_{3}^{-}}{{H}_{2} C {O}_{3}}\right)$

$1.085 = \log \left(\frac{H C {O}_{3}^{-}}{{H}_{2} C {O}_{3}}\right)$

$\left(\frac{H C {O}_{3}^{-}}{{H}_{2} C {O}_{3}}\right)$ = ${10}^{1.085} = 12.162$

$\left(\frac{{H}_{2} C {O}_{3}}{H C {O}_{3}^{-}}\right)$ = $\left(\frac{1}{12.162}\right) = 0.0822$

$\left[{H}_{2} C {O}_{3}\right] : \left[H C {O}_{3}^{-}\right] = 0.0822 : 1.0000$