How is the Henderson-Hasselbalch equation used to calculate the ratio of #H_2CO_3# to #HCO_3^-# in blood having a pH of 7.40?

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Answer 1 · 2017-05-29T03:16:46

#([HCO_3^-]:[H_2CO_3]) = 10:1#

#pH = pKa + log_10([[HCO_3^-]]/[[H_2CO_3]])#

#Ka_1(H_2CO_3) = 4.5x10^(-7)#
=> #pKa = -logK_a# = #-log(4.5x10^(-7))# =#6.4#

Given #pH = 7.4#

Substitute into HH Equation and solve for #([[HCO_3^-]]/[[H_2CO_3]])#

#7.4 = 6.4 + log_10([[HCO_3^-]]/[[H_2CO_3]])#

#log_10([[HCO_3^-]]/[[H_2CO_3]]) = 7.4 - 6.4 = 1.0#

#([[HCO_3^-]]/[[H_2CO_3]]) = 10^1.0 = 10#

=> #([HCO_3^-]:[H_2CO_3]) = 10:1#