The pH of blood is 7-4- What is the ratio of - HPO 4 2- - - H 2 PO 4 - - in the blood- pK a H 2 PO 4 - -7-1-

The correct option is A 2 : 1 pH= pK _ a +log[ [ Salt ] #160; [ Acid ] #160; #160; ] #160; pH=7.4 pK_a=7.1 ⟹ 7.4=7.1+log[ [ HPO _ ...

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Byju's Answer

Standard VIII

Chemistry

Ways to Define Concentration

The pH of blo...

Question

The pH of blood is 7.4. What is the ratio of $\frac{[{H P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}$ in the blood. ${p K}_{a} (H_{2} {P O}_{4}^{-}) = 7.1$ ?

2 : 1

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Solution

The correct option is A 2 : 1
$p H = {p K}_{a} + log [\frac{[S a l t]}{[A c i d]}]$
$p H = 7.4$
$p K_{a} = 7.1$
$⟹ 7.4 = 7.1 + log ⎡ ⎢ ⎣ \frac{[{H P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]} ⎤ ⎥ ⎦ ⟹ log ⎡ ⎢ ⎣ \frac{[{H P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]} ⎤ ⎥ ⎦ = 0.3 = log 2$
$∴$ Ratio of the concentration of $H P O_{4}^{2 -}$ and $H_{2} P O_{4}^{-}$ is $2 : 1$

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Similar questions

Q. In a buffer solution consisting

N a H_{2} {P O}_{4}

and

{N a}_{2} H {P O}_{4}

(1)

N a H_{2} {P O}_{4}

is acid and

{N a}_{2} H {P O}_{4}

is salt
(2) The pH of solution can be calculated using the relation:

p H = p K_{2} + {log}_{10} \frac{[H {P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}

(3) The

{N a}_{2} H {P O}_{4}

is acid and

N a H_{2} {P O}_{4}

is salt
(4) The pH cannot be calculated.

Q. Three reactions involving

H_{2} {P O}_{4}^{-}

are given below:
(i)

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

(ii)

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

(iii)

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

Hence, pH of

0.01 M

N a H_{2} {P O}_{4}

is:

K_{a_{1}}, K_{a_{2}}, K_{a_{3}}

values for

H_{3} {P O}_{4}

are

10^{- 3}, 10^{- 8}, 10^{- 12}

respectively. If

K_{w} (H_{2} O) = 10^{- 14}

, then
(i) What is dissociation constant of

H {P O}_{4}^{2 -}

?
(ii) What is

K_{b}

H {P O}_{4}^{2 -}

(iii) What is

K_{b}

H_{2} {P O}_{4}^{2 -}

?
(iv) What order of

K_{b}

{P O}_{4}^{3 -} (K_{b_{3}}), H {P O}_{4}^{2 -} (K_{b_{2}})

and

H_{2} {P O}_{4}^{-} (K_{b_{1}})

Q. Calculate

[H^{+}]

[H_{2} {P O}_{4}]

[H P O_{4}^{2 -}]

and

[{P O}_{4}^{3 -}]

in a

0.01 M

solution of

H_{3} {P O}_{4}

.
Take

K_{1} = 7.225 \times 10^{- 3}

K_{2} = 6.8 \times 10^{- 8}

K_{3} = 4.5 \times 10^{- 13}

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The pH of blood is 7.4. What is the ratio of [HPO2−4][H2PO−4] in the blood. pKa(H2PO−4)=7.1?

The correct option is A 2 : 1pH=pKa+log[[Salt][Acid]]pH=7.4pKa=7.1⟹7.4=7.1+log⎡⎢⎣[HPO2−4][H2PO−4]⎤⎥⎦⟹log⎡⎢⎣[HPO2−4][H2PO−4]⎤⎥⎦=0.3=log2∴ Ratio of the concentration of HPO2−4 and H2PO−4 is 2:1

The pH of blood is 7.4. What is the ratio of $\frac{[{H P O}_{4}^{2 -}]}{[H_{2} {P O}_{4}^{-}]}$ in the blood. ${p K}_{a} (H_{2} {P O}_{4}^{-}) = 7.1$ ?