Question #dc4c4

2 Answers
Apr 29, 2017

Answer:

#[[H_2CO_3]]/[[HCO_3^-]]=0.0823#

Explanation:

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

#H_2CO_3rightleftharpoonsH^++HCO_3^-#

Remember the dissociation constant will be the ratio of the dissociated products (#H^+# and #HCO_3^-#) to the carbonic acid (#H_2CO_3#).

#K_d=([H^+][HCO_3^-])/[[H_2CO_3]]#

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

#([HCO_3^-])/[[H_2CO_3]]=K_d/[[H^+]]#

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

#([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d#

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

#pH=-log[H^+]#
#[H^+]=10^(-pH)#
#[H^+]=10^-7.45#
#[H^+]=3.55 *10^-8M#

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

#([H_2CO_3])/[[HCO_3^-]]=[[H^+]]/K_d#
#([H_2CO_3])/[[HCO_3^-]]=(3.55*10^-8)/(4.31*10^-7)#
#([H_2CO_3])/[[HCO_3^-]]=0.0823#

Apr 29, 2017

Answer:

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

Explanation:

#pH = pKa + log([HCO_3^-]/[H_2CO_3])#

#7.45 = -log(4.31E-7)# + #log([HCO_3^-]/[H_2CO_3])#

#7.45 = 6.37 + log([HCO_3^-]/[H_2CO_3])#

#1.085 = log([HCO_3^-]/[H_2CO_3])#

#([HCO_3^-]/[H_2CO_3])# = #10^(1.085) = 12.162#

#([H_2CO_3]/[HCO_3^-])# = #(1/12.162) = 0.0822#

#[H_2CO_3]:[HCO_3^-] = 0.0822 : 1.0000#