## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.6 - Page 670: d

#### Answer

$pH = 10.70$

#### Work Step by Step

1000ml = 1L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.1* 0.05 = 5 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.1* 0.0505 = 5.2 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - Total volume: 0.05 + 0.0505 = 0.1005L 3. Since the acid is the limiting reactant, only $0.005$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.005 - 0.005 = 0M$. $[NaOH] = 0.00505 - 0.005 = 5 \times 10^{-5}$ mol Concentration: $\frac{5.0 \times 10^{-5}}{ 0.1005} = 5 \times 10^{-4}M$ - The only significant electrolyte in the solution is $NaOH$, which is a strong base, so: $[OH^-] = [NaOH] = 5.0 \times 10^{-4}M$ 4. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 5.0 \times 10^{- 4})$ $pOH = 3.30$ 5. Find the pH: $pH + pOH = 14$ $pH + 3.3 = 14$ $pH = 10.70$

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