± Composition of Inorganic Conjugate Acid-Base Pairs in Blood pH of blood The tw
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Question
± Composition of Inorganic Conjugate Acid-Base Pairs in Blood
pH of blood
The two most important inorganic blood buffers (acid-base conjugate pairs) are the phosphoric acid and carbonic acid systems. The pH of normal blood is 7.40.
Part A
At a pH of 7.40, what is the ratio of the molar concentrations of PO43 to HPO42?
Express your answer using two significant figures.
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Part B
At a pH of 7.40, what is the ratio of the molar concentrations of HPO42 to H2PO4?
Express your answer using two significant figures.
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Part C
At a pH of 7.40, what is the ratio of the molar concentrations of H2PO4 to H3PO4?
Express your answer using two significant figures.
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Part D
At a pH of 7.40, what is the ratio of the molar concentrations of CO32 to HCO3?
Express your answer using two significant figures.
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Part E
At a pH of 7.40, what is the ratio of the molar concentrations of HCO3 to H2CO3?
Express your answer using two significant figures.
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Part F
Which conjugate acid-base pair is most responsible for maintaining the pH of blood?
Which conjugate acid-base pair is most responsible for maintaining the of blood?
[CO32][HCO3]
[HCO3][H2CO3]
[H2PO4][H3PO4]
[PO43][HPO42]
[HPO42][H2PO4]
± Composition of Inorganic Conjugate Acid-Base Pairs in Blood
The following table gives the acid-dissociation constant values, Ka, for carbonic acid, H2CO3, and phosphoric acid, H3PO4. Acid Ka1 Ka2 Ka3 H2CO3 4.3×107 5.6×1011 not applicable H3PO4 7.2×103 6.2×108 4.2×1013pH of blood
The two most important inorganic blood buffers (acid-base conjugate pairs) are the phosphoric acid and carbonic acid systems. The pH of normal blood is 7.40.
Part A
At a pH of 7.40, what is the ratio of the molar concentrations of PO43 to HPO42?
Express your answer using two significant figures.
[PO43][HPO42] =SubmitHintsMy AnswersGive UpReview Part
Part B
At a pH of 7.40, what is the ratio of the molar concentrations of HPO42 to H2PO4?
Express your answer using two significant figures.
[HPO42][H2PO4] =SubmitHintsMy AnswersGive UpReview Part
Part C
At a pH of 7.40, what is the ratio of the molar concentrations of H2PO4 to H3PO4?
Express your answer using two significant figures.
[H2PO4][H3PO4] =SubmitHintsMy AnswersGive UpReview Part
Part D
At a pH of 7.40, what is the ratio of the molar concentrations of CO32 to HCO3?
Express your answer using two significant figures.
[CO32][HCO3] =SubmitHintsMy AnswersGive UpReview Part
Part E
At a pH of 7.40, what is the ratio of the molar concentrations of HCO3 to H2CO3?
Express your answer using two significant figures.
[HCO3][H2CO3] =SubmitHintsMy AnswersGive UpReview Part
Part F
Normal metabolic processes produce significant amounts of acid, sometimes up to 10 mol/day, in the human body. Thus it is desirable that the buffer system for blood contain more conjugate base than acid.Which conjugate acid-base pair is most responsible for maintaining the pH of blood?
Normal metabolic processes produce significant amounts of acid, sometimes up to 10 , in the human body. Thus it is desirable that the buffer system for blood contain more conjugate base than acid.Which conjugate acid-base pair is most responsible for maintaining the of blood?
[CO32][HCO3]
[HCO3][H2CO3]
[H2PO4][H3PO4]
[PO43][HPO42]
[HPO42][H2PO4]
Explanation / Answer
Part-A
PH = PKa + log[PO43-]/[HPO42-]
Pka = -logKa
= -log4.2*10-13
= 12.3767
PH = PKa + log[PO43-]/[HPO42-]
7.4 =12.3767 + log[PO43-]/[HPO42-]
7.4-12.3767 = log[PO43-]/[HPO42-]
log[PO43-]/[HPO42-] = -4.9767
[PO43-]/[HPO42-] = 10-4.9767 = 1.05*10-5
Part-B
PH = PKa + log[HPO42-]/[H2PO4-]
PKa = -logKa
= -log6.2*10-8
= 7.207
7.4 = 7.207 + log[PO43-]/[HPO4-]
7.4-7.207 = log[HPO42-]/[H2PO4-]
log[HPO42-]/[H2PO4-] = 0.193
[HPO42-]/[H2PO4-] = 100.193 = 1.6
part-C
PH = Pka + log[H2PO4-]/[H3PO4]
PKa = -logKa
= -log7.2*10-3
= 2.1426
PH = Pka + log[H2PO4-]/[H3PO4]
7.4 = 2.1426 + log[H2PO4-]/[H3PO4]
log[H2PO4-]/[H3PO4] = 7.4-2.1426
log[H2PO4-]/[H3PO4] = 5.2574
[H2PO4-]/[H3PO4] = 105.2574
[H2PO4-]/[H3PO4] = 180883/1
Part-D
PH = Pka + log[CO32-]/[HCO3-]
Pka = -logKa
= -log5.6*10-11
= 10.2518
PH = Pka + log[CO32-]/[HCO3-]
7.4 = 10.2518+ log[CO32-]/[HCO3-]
log[CO32-]/[HCO3-] = 7.4-10.2518
log[CO32-]/[HCO3-] = -2.8518
[CO32-]/[HCO3-] = 10-2.8518
[CO32-]/[HCO3-] = 0.0014
part-e
PH = PKa + log[HCO3-]/[H2CO3]
PKa = -logKa
= -log4.3*10-7
= 6.3665
PH = PKa + log[HCO3-]/[H2CO3]
7.4 = 6.3665 + log[HCO3-]/[H2CO3]
log[HCO3-]/[H2CO3] = 7.4-6.3665
log[HCO3-]/[H2CO3] = 1.0335
[HCO3-]/[H2CO3] = 101.0335
[HCO3-]/[H2CO3] = 10.80
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