Question

(a) carbonate buffers are important in regulating the ph of blood at 7.40. what is the concentration ratio of co2 (usually written h2co3) to hco3- in blood at ph = 7.40? h2co3(aq) equilibrium reaction arrow hco3-(aq) + h+(aq) ka = 4.3 ✕ 10-7 webassign will check your answer for the correct number of significant figures. (b) phosphate buffers are important in regulating the ph of intracellular fluids at ph values generally between 7.1 and 7.2. what is the concentration ratio of h2po4- to hpo42- in intracellular fluid at ph = 7.15? h2po4-(aq) equilibrium reaction arrow hpo42-(aq) + h+(aq) ka = 6.2 ✕ 10-8 webassign will check your answer for the correct number of significant figures. (c) why is a buffer composed of h3po4 and h2po4- ineffective in buffering the ph of intracellular fluid? h3po4(aq) equilibrium reaction arrow h2po4-(aq) + h+(aq) ka = 7.5 ✕ 10-3

Answer 1

A) when the balanced equation of the reaction is:
H2CO3(aq) → HCO3 -(aq) + (H+)
and when we have Ka = 4.3 x10^-7 & PH = 7.4 
So first we will get PKa = -㏒ Ka
PKa = -㏒(4.3x10^-7) = 6.37 by substitution with Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
PH= PKa + ㏒[HCO3-]/[H2CO3]
㏒[HCO3-]/[H2CO3] = PH-Pka
[HCO3-] /[H2CO3] = 10^(7.4 - 6.37)
∴[HCO3-]/[H2CO3] = 11.7
∴[H2CO3]/[HCO3-] = 1/11.7 =  0.09

B) when The balanced equation for this reaction is:
H2PO42-(aq) → HPO4-(aq) + H+
and when we have Ka = 6.2x10^-8 & PH = 7.15
So Pka= -㏒Ka = -㏒(6.2x10^-8) = 7.2 by substitution by Pka value in the following formula:
PH = Pka + ㏒[salt/acid]
7.15= 7.2 + ㏒[HPO4]/[H2PO4] 
-0.05 = ㏒[HPO4]/[H2PO4]
∴[HPO4]/[H2PO4] = 10^-0.05 = 0.89
∴[H2PO4]/[HPO4] = 1/0.89 = 1.12

c) H3PO4(aq) ↔ H2PO-(aq) + H+
the answer is: because we have Ka =7.5x10^-3 and it is a high value of Ka to make a good buffer, also we need a week acid with th salt of the week acid as H3PO4 is a strong acid so it does'nt make a goof buffer.