## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 17 Principles of Chemical Reactivity: Other Aspects of Aqueous Equilibria - Study Questions - Page 677a: 24

#### Answer

(a) $pH = 7.21$ (b) $pH = 7.24$

#### Work Step by Step

1. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 6.2 \times 10^{- 8})$ $pKa = 7.21$ 2. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.132}{0.132}$ - 1: It is. 3. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.13}{6.2 \times 10^{-8}} = 2.1\times 10^{6}$ - $\frac{0.13}{6.2 \times 10^{-8}} = 2.1\times 10^{6}$ 4. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 7.21 + log(\frac{0.132}{0.132})$ $pH = 7.21 + log(1)$ $pH = 7.21 + 0$ $pH = 7.21$ 5. Calculate the molar mass $(NaOH)$: 22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol 6. Calculate the number of moles $(NaOH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.425}{ 40}$ $n(moles) = 0.0106$ 7. Find the concentration in mol/L $(NaOH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.0106}{ 2}$ $C(mol/L) = 5.3\times 10^{- 3}$ 8. The addition of a strong base will reduce the acid concentration, and increase the base concentration by its own value: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 7.21 + log(\frac{[0.132 + 5.3 \times 10^{-3}]}{[0.132 - 5.3 \times 10^{-3}]})$ $pH = 7.21 + log(\frac{0.137}{[0.127]})$ $pH = 7.21 +log(1.08)$ $pH = 7.21 + 0.03$ $pH = 7.24$

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