Answer:
a) H2CO3 (aq) HCO3-(aq) + H+ (aq)
Acid Conjugate base
pH of the buffer is given by Henderson-Hasselbalch equation as,
pH = pKa + log([Conjugate base]/[Acid])
For H2CO3 and HCO3- buffer,
pH = pKa (H2CO3) + log([HCO3-]/[H2CO3]) ..... (1)
Given data: pH of blood = 7.40,
Ka = 4.3*10-7. pKa = -logKa = -log(4.3*10-7) = 6.37
Placing these values in eq.(1)
7.40 = 6.37 + log([HCO3-]/[H2CO3])
log([HCO3-]/[H2CO3]) = 7.40-6.37
log([HCO3-]/[H2CO3]) = 1.03
[HCO3-]/[H2CO3] = 101.03.
[HCO3-]/[H2CO3] = 10.7
On reciprocal,
[H2CO3] / [HCO3-] = 1 / 10.7
[H2CO3] / [HCO3-] = 0.09
Is the required ratio.
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b) H2PO4- (aq) HPO42- (aq) + H+ (aq)
Acid Conjugate base
pH of the buffer is given by Henderson-Hasselbalch equation as,
pH = pKa + log([Conjugate base]/[Acid])
For H2PO4- (aq) and HPO42- (aq)
pH = pKa (H2CO3) + log([HPO42-]/[H2PO4-]) ..... (1)
Given data: pH of blood = 7.15,
Ka = 6.2*10-8. pKa = -logKa = -log(6.2*10-8) = 7.21
Placing these values in eq.(1)
7.15 = 7.21 + log([HPO42-]/[H2PO4-])
log([HPO42-]/[H2PO4-]) = 7.15-7.21
log([HPO42-]/[H2PO4-]) = -0.06
[HPO42-]/[H2PO4-] = 10-0.06.
[HPO42-]/[H2PO4-] = 0.87
On reciprocal,
[H2PO42-]/[HPO42-] = 1/0.87
[H2PO42-]/[HPO42-] = 1.15
Is the required ratio.
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