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Standard XII

Chemistry

Properties of Water

The [ HCO 3-]...

Question

The $\frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$ ratio in blood (pH = 7.4) is approximately (pKa = 6.1 for $H_{2} C O_{3})$ :
(Take log 2 = 0.3, log 3 = 0.5, log 7 = 0.85)

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Solution

The correct option is A 20
$p H = p K a + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or,
$7.4 = 6.1 + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or,
$\frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]} = 20$

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Similar questions

Q. The

[H C O_{3}^{-}] / [H_{2} C O_{3}]

ratio in the blood (pH = 7.4) is approximately

(p K_{a} = 6.1 f o r H_{2} C O_{3})

log 20= 1.301

The pH of blood is 7.4. If the buffer in bloods is $C O_{2} a n d H C O_{3}^{-}$ , calculate the ratio of conjugate base to acid $(H_{2} C O_{3})$ to maintain the pH of the blood. Given $K_{1}$ of $H_{2} C O_{3} = 4.5 \times 10^{- 7}$

Q. The

p H

of a blood stream is maintained by a proper balance of

H_{2} C O_{3}

and

N a H C O_{3} .

What volume of 5 M

N a H C O_{3}

solution should be mixed with 10 mL of a sample of solution which is 2.5 M in

H_{2} C O_{3},

in order to maintain a

p H = 7.4.

(Take

p K_{a_{1}}

for

H_{2} C O_{3} = 6.7, l o g 2 = 0.3, l o g 5 = 0.7)

p K_{a_{1}}

of carbonic acid in blood at body temperature (

37^{o} C

) is 6.1. Hence, ratio

[H C O_{3}^{-}] / [H_{2} C O_{3}]

is approximately:

Q. The

p H

of a blood stream is maintained by a proper balance of

H_{2} C O_{3}

and

N a H C O_{3} .

What volume of

2.5 M

H_{2} C O_{3},

solution should be mixed with

25 mL

of a sample of solution which is

5 M

N a H C O_{3}

in order to maintain a

p H = 7.4.

(Take

p K_{a_{1}}

for

H_{2} C O_{3} = 6.7, l o g 2 = 0.3, l o g 5 = 0.7)

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The [HCO−3][H2CO3] ratio in blood (pH = 7.4) is approximately (pKa = 6.1 for H2CO3): (Take log 2 = 0.3, log 3 = 0.5, log 7 = 0.85)

The correct option is A 20pH=pKa+log[HCO−3][H2CO3] or, 7.4=6.1+log[HCO−3][H2CO3] or, [HCO−3][H2CO3]=20

The $\frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$ ratio in blood (pH = 7.4) is approximately (pKa = 6.1 for $H_{2} C O_{3})$ :
(Take log 2 = 0.3, log 3 = 0.5, log 7 = 0.85)

The correct option is A 20
$p H = p K a + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or,
$7.4 = 6.1 + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or,
$\frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]} = 20$