The pH of blood stream is maintained by a proper balance of $H_{2} {CO}_{3} {and NaHCO}_{3}$ concentration. The volume of ${10M NaHCO}_{3}$ solution which is mixed with 10 mL sample of blood which is 4 M in $H_{2} {CO}_{3}$ in order to maintain a pH of 7.4 is x mL. $K_{a} {for H}_{2} {CO}_{3}$ in blood is $4 \times 10^{- 7}$ . What will be the value of $\frac{x}{10}$ ?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 4.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$[H_{2} {CO}_{3}]$ in blood = 4M $[{NaHCO}_{3}] =10M$
Volume of blood = 10 mL
Let the volume of ${NaHCO}_{3}$ used = x mL
$[H_{2} {CO}_{3}]$ in mixture $= \frac{4×10}{(x+10)}$
$[{NaHCO}_{3}] in mixture= \frac{10×x}{(x+10)} {pH=pK}_{a} +log \frac{salt}{acid} 7.4=7-log4 \frac{(10 \times x) (x + 10)}{(x + 10) (4 \times 10)} \Rightarrow 7.4 = 6.4 + \log \frac{10 x}{40} log \frac{10x}{40} =1 so \frac{10x}{40} =10$
Hence $\frac{x}{10} = 4$