Calculate the pH of normal arterial blood using the information given below: PK\

Question

Calculate the pH of normal arterial blood using the information given below: PK' = 6.1 for CO_2 doubleheadarrow H^+ + HC0_3^- [CO_2 + H2O doubleheadarrow H_2CO_3 doubleheadarrow H^+ + HCO_3^- with pKa = 6.35 for second equilibrium. When both K_eq values are combined the resulting pK' = 6.1 Use this value answer this question so [HA] = [CO_2} and [A] = [HCO_3^-] in the buffering system in the blood. In arterial blood, [HC0_3^-] = 24 mM; [C0_2] = 1.2 mM Suppose the normal arterial blood is suddenly made to accept 9 mM of HCI(aq) i.e. H^+ ions What is the resulting pH of the blood if no CO_2 is allowed to escape? What is the resulting pH of the blood if 9 mM of CO_2 can be quickly exhaled by normal processes? Under these conditions, a person would actually begin to hyperventilate. How will this help to restore normal pH?

Explanation / Answer

a) Given:

CO2 ----- H+ + HCO3- --------(1) pK' = 6.1 , K' = 10^-pK' = 10^-6.1 = 7.94 * 10^-7

H2CO3 ------ H+ + HCO3- -----(2) pKa = 6.35, Ka = 10^-pka = 10^-6.35 = 4.47 8 10^-7

We know that: pKa = -[log Ka]

Effective , pKe for the aboce equations = -log[pK'.pKa] = -log[7.94*10^-7 * 4.47 * 10^-7] = 12.45

As per Henderson- Hasselbach equation:

pH = pKe + log [HCO3^-]/[CO2] = 12.45 + log[24/1.2] = 12.45 + 1.30 = 13.75

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