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 _ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 : 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 : 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 : 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $2 : 1$

Suggest Corrections

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}

Join BYJU'S Learning Program

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. ${p K}_{a} (H_{2} {P O}_{4}^{-}) = 7.1$ ?