SATHEE: Chemical and Ionic Equilibrium 2 Question 65

Chemical and Ionic Equilibrium 2 Question 65

65. The $pH$ of blood stream is maintained by a proper balance of $\mathrm{H}{2} \mathrm{CO}{3} $a n d$ \mathrm{NaHCO}{3} $c o n c e n t r a t i o n s . W h a t v o l u m e o f$ 5 \mathrm{M}\mathrm{NaHCO}{3} $s o l u t i o n s h o u l d b e m i x e d w i t h a$ 10 \mathrm{~mL} $s a m p l e o f b l o o d w h i c h i s$ 2 \mathrm{M} $i n$ \mathrm{H}{2} \mathrm{CO}{3} $, i n o r d e r t o m a i n t a i n a$ \mathrm{pH} $o f 7.4 ?$ \left(K_{a}\right. $f o r$ \mathrm{H}{2} \mathrm{CO}{3} $i n b l o o d i s$ 7.8 \times 10^{-7}$ )

(1993, 2M)

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Solution:

Mixing $\mathrm{H}{2} \mathrm{CO}{3} $w i t h$ \mathrm{NaHCO}_{3}$ results in buffer solution.

$\mathrm{pH}=\mathrm{p} K_{a}+\log \frac{\left[\mathrm{NaHCO}{3}\right]}{\left[\mathrm{H}{2} \mathrm{CO}{3}\right]}=\mathrm{p} K{a}+\log \frac{n\left(\mathrm{NaHCO}{3}\right)}{n\left(\mathrm{H}{2} \mathrm{CO}_{3}\right)}$

$\Rightarrow 7.4 = - \log (7.8 \times 10^{- 7}) + \log \frac{x}{20}$

$\Rightarrow x = 400 mmol$

${NaHCO}_{3} = 5 \times V \Rightarrow V = 80 mL$

Chemical and Ionic Equilibrium 2 Question 65

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Chemical and Ionic Equilibrium 2 Question 65

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

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