Question #dc4c4

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Question #dc4c4

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Answer 1 · 2017-04-29T03:09:41

$\frac{[H_{2} C O_{3}]}{[H C O_{3}^{-}]} = 0.0823$ $[[H_2CO_3]]/[[HCO_3^-]]=0.0823$

Explanation:

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

$H_{2} C O_{3} ⇌ H^{+} + H C O_{3}^{-}$ $H_2CO_3rightleftharpoonsH^++HCO_3^-$

Remember the dissociation constant will be the ratio of the dissociated products ( $H^{+}$ $H^+$ and $H C O_{3}^{-}$ $HCO_3^-$ ) to the carbonic acid ( $H_{2} C O_{3}$ $H_2CO_3$ ).

$K_{d} = \frac{[H^{+}] [H C O_{3}^{-}]}{[H_{2} C O_{3}]}$ $K_d=([H^+][HCO_3^-])/[[H_2CO_3]]$

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

$\frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]} = \frac{K_{d}}{[H^{+}]}$ $([HCO_3^-])/[[H_2CO_3]]=K_d/[[H^+]]$

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

$\frac{[H_{2} C O_{3}]}{[H C O_{3}^{-}]} = \frac{[H^{+}]}{K_{d}}$ $([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d$

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

$p H = - log [H^{+}]$ $pH=-log[H^+]$
$[H^{+}] = 10^{- p H}$ $[H^+]=10^(-pH)$
$[H^{+}] = 10^{- 7.45}$ $[H^+]=10^-7.45$
$[H^{+}] = 3.55 \cdot 10^{- 8} M$ $[H^+]=3.55 *10^-8M$

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

$\frac{[H_{2} C O_{3}]}{[H C O_{3}^{-}]} = \frac{[H^{+}]}{K_{d}}$ $([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d$
$\frac{[H_{2} C O_{3}]}{[H C O_{3}^{-}]} = \frac{3.55 \cdot 10^{- 8}}{4.31 \cdot 10^{- 7}}$ $([H_2CO_3])/[[HCO_3^-]]=(3.55*10^-8)/(4.31*10^-7)$
$\frac{[H_{2} C O_{3}]}{[H C O_{3}^{-}]} = 0.0823$ $([H_2CO_3])/[[HCO_3^-]]=0.0823$

Answer 2 · 2017-04-29T03:32:24

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

Explanation:

$p H = p K a + log (\frac{H C O_{3}^{-}}{H_{2} C O_{3}})$ $pH = pKa + log([HCO_3^-]/[H_2CO_3])$

$7.45 = - log (4.31 E - 7)$ $7.45 = -log(4.31E-7)$ + $log (\frac{H C O_{3}^{-}}{H_{2} C O_{3}})$ $log([HCO_3^-]/[H_2CO_3])$

$7.45 = 6.37 + log (\frac{H C O_{3}^{-}}{H_{2} C O_{3}})$ $7.45 = 6.37 + log([HCO_3^-]/[H_2CO_3])$

$1.085 = log (\frac{H C O_{3}^{-}}{H_{2} C O_{3}})$ $1.085 = log([HCO_3^-]/[H_2CO_3])$

$(\frac{H C O_{3}^{-}}{H_{2} C O_{3}})$ $([HCO_3^-]/[H_2CO_3])$ = $10^{1.085} = 12.162$ $10^(1.085) = 12.162$

$(\frac{H_{2} C O_{3}}{H C O_{3}^{-}})$ $([H_2CO_3]/[HCO_3^-])$ = $(\frac{1}{12.162}) = 0.0822$ $(1/12.162) = 0.0822$

$[H_{2} C O_{3}] : [H C O_{3}^{-}] = 0.0822 : 1.0000$ $[H_2CO_3]:[HCO_3^-] = 0.0822 : 1.0000$

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