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Henderson Hassleback Equation

The [ HCO 3-]...

Question

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

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Solution

The correct option is A 20
From Henderson equation,
$p H = p K_{a} + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or, $7.4 = 6.1 + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
$1.3 = l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
$\Rightarrow \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]} = 20$

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Similar questions

Q. 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 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 $H_{2} C O_{3} = 4.5 \times 10^{- 7}$

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. Blood

p H

is controlled by the concentrations of

H_{2} C O_{3}

and

H C O_{3}^{-}

. In the presence of

N a H C O_{3}

p H

of blood:

Q. In blood (

p H = 7.4

), aspirin has

p K_{a}

of 3.4. What is the ratio of

A^{-}

H A

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Henderson Hassleback Equation

Standard XII Chemistry

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The [HCO−3]/[H2CO3] ratio in the blood (pH = 7.4) is approximately (pKa=6.1 for H2CO3) log 20= 1.301

The correct option is A 20From Henderson equation, pH=pKa+log[HCO−3][H2CO3] or, 7.4=6.1+log[HCO−3][H2CO3] 1.3=log[HCO−3][H2CO3] ⇒[HCO−3][H2CO3]=20

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 correct option is A 20
From Henderson equation,
$p H = p K_{a} + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
or, $7.4 = 6.1 + l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
$1.3 = l o g \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]}$
$\Rightarrow \frac{[H C O_{3}^{-}]}{[H_{2} C O_{3}]} = 20$