Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.7 - Page 673: d

Answer

$pH = 11.00$

Work Step by Step

- Find the numbers of moles: $C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.05 = 5 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.1* 0.051 = 5.1 \times 10^{-3}$ moles Write the acid-base reaction: $CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$ - Total volume: 0.05 + 0.051 = 0.101L Since the acid is the limiting reactant, only $ 0.005$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[CH_3COOH] = 0.005 - 0.005 = 0M$. $[NaOH] = 0.0051 - 0.005 = 1.0 \times 10^{-4}$ mol Concentration: $\frac{1.0 \times 10^{-4}}{ 0.101} = 9.901 \times 10^{-4}M$ $[NaCH_3COO] = 0 + 0.005 = 0.005$ moles. Concentration: $\frac{ 0.005}{ 0.101} = 0.0495M$ - We have a strong and a weak base. We can ignore the weak one, and we calculate the pH based only on the strong base concentration: $[OH^-] = [NaOH]$ $pOH = -log[OH^-]$ $pOH = -log( 9.901 \times 10^{- 4})$ $pOH = 3.004$ $pH + pOH = 14$ $pH + 3.004 = 14$ $pH = 11.00$
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