The pH of blood is maintained by a proper bal | vedclass

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(C) Let the volume of $NaHCO_3$ solution be $x \, mL$. Moles of $NaHCO_3$ in $x \, mL$ of $5 \, M$ solution $= \frac{5x}{1000} = 0.005x \, mol$. Moles

Vedclass · Accepted Answer

$78.36$

Vedclass · Answer

(C) Let the volume of $NaHCO_3$ solution be $x \, mL$. Moles of $NaHCO_3$ in $x \, mL$ of $5 \, M$ solution $= \frac{5x}{1000} = 0.005x \, mol$. Moles of $H_2CO_3$ in $10 \, mL$ of $2 \, M$ solution $= \frac{2 	imes 10}{1000} = 0.02 \, mol$. Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$ $pK_a = -\log(7.8 	imes 10^{-7}) = 7 - \log(7.8) = 7 - 0.892 = 6.108$. $7.4 = 6.108 + \log \left( \frac{0.005x}{0.02} ight)$. $7.4 - 6.108 = \log(0.25x)$. $1.292 = \log(0.25x)$. $0.25x = 10^{1.292} \approx 19.59$. $x = \frac{19.59}{0.25} = 78.36 \, mL$.

The $pH$ of blood is maintained by a proper balance of $H_2CO_3$ and $NaHCO_3$ concentrations. What volume of $5 \, M \, NaHCO_3$ solution should be mixed with $10 \, mL$ of $2 \, M \, H_2CO_3$ to maintain a $pH$ of $7.4$ (in $, mL$)? (Given: $K_a$ for $H_2CO_3 = 7.8 \times 10^{-7}$)

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