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 $\frac{x}{10}$ ?

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answer is 4.

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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 $\frac{x}{10} = 4$