In blood pH -7-4- aspirin has pK a of 3-4- What is the ratio of A -to HA-A- 1- 10000B- 1000- 1C- 10000- 1D- 100- 1

The correct option is C 10000:1Given, nbsp;pK_a = 3.4, pH = 7.4 Using nbsp;Henderson Hasselbalch equation: nbsp; pH = pK_a + log ([conjugate base] [aci ...

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Standard XII

Chemistry

Acidic Buffer Action

In blood pH =...

Question

In blood ( $p H = 7.4$ ), aspirin has $p K_{a}$ of 3.4. What is the ratio of $A^{-}$ to $H A$ ?

1:10000

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1000:1

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10000:1

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100:1

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Solution

The correct option is C 10000:1
Given, $p K_{a} = 3.4, p H = 7.4$
Using Henderson-Hasselbalch equation:
$p H = p K_{a} + l o g (\frac{[conjugate base]}{[acid]}$
$p H = p K_{a} + l o g (\frac{[A^{-}]}{[H A]}$
$7.4 = 3.4 + l o g (\frac{[A^{-}]}{[H A])}$
$7.4 = 3.4 = l o g (\frac{[A^{-}]}{[H A]}$
$4 = l o g (\frac{[A^{-}]}{[H A]}$
$10^{4} = \frac{[A^{-}]}{[H A]}$
$[A^{-}] : [H A] = 10000 : 1$

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Acidic Buffer Action

Standard XII Chemistry

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In blood (pH=7.4), aspirin has pKa of 3.4. What is the ratio of A− to HA?

The correct option is C 10000:1Given, pKa=3.4, pH=7.4 Using Henderson-Hasselbalch equation: pH=pKa+log([conjugate base][acid] pH=pKa+log([A−][HA] 7.4=3.4+log([A−][HA]) 7.4=3.4=log([A−][HA] 4=log([A−][HA] 104=[A−][HA] [A−]:[HA]=10000:1

In blood ( $p H = 7.4$ ), aspirin has $p K_{a}$ of 3.4. What is the ratio of $A^{-}$ to $H A$ ?