# Chapter 15 - Additional Aqueous Equilibria - Exercise 15.2 - Blood pH - Page 658: a

$pH = 7.423$

#### Work Step by Step

1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 6.3 \times 10^{- 8})$ $pKa = 7.201$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{2.5 \times 10^{-3}}{1.5 \times 10^{-3}} = 1.667$ : It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{2.5 \times 10^{-3}}{ 6.3 \times 10^{-8}} = 3.968\times 10^{4}$ - $\frac{1.5 \times 10^{-3}}{ 6.3 \times 10^{-8}} = 2.381\times 10^{4}$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 7.201 + log(\frac{2.5 \times 10^{-3}}{1.5 \times 10^{-3}})$ $pH = 7.201 + 0.2218$ $pH = 7.423$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.