人体血液里最主要的缓冲体系是碳酸氢盐缓冲体系(H
2CO
3/HCO
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsubsup%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmath%3E)
),该体系的主要作用是维持血液中pH保持稳定。在人体正常体温时,H
2CO
3![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmtext%3E%E2%87%8C%3C%2Fmtext%3E%3C%2Fmath%3E)
HCO
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsubsup%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmath%3E)
+H
+的K
a=10
−6.1 , 正常人的血液中c(HCO
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsubsup%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmo%3E%E2%88%92%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmath%3E)
):c(H
2CO
3)≈20:1,则下列判断正确的是(已知lg2=0.3)( )
- A、正常人血液内Kw=10-14
- B、当过量的碱进入血液中时,只发生反应HCO
+OH-=CO
+H2O - C、正常人血液中存在:c(HCO
)+c(OH-)+2c(CO
)=c(H+)+c(H2CO3) - D、由题给数据可算得正常人血液的pH约为7.4