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

$pH = 2.28$

#### 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 $C(NaOH) * V(NaOH) = 0.1* 0.045 = 4.5 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - Total volume: 0.05 + 0.045 = 0.095L 3. Since the base is the limiting reactant, only $0.0045$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.005 - 0.0045 = 5 \times 10^{-4}$ moles. Concentration: $\frac{5 \times 10^{-4}}{ 0.095} = 5.3 \times 10^{-3}M$ $[NaOH] = 0.0045 - 0.0045 = 0$ moles - Since $HCl$ is a strong acid: $[HCl] = [H_3O^+] = 5.3 \times 10^{-3}M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 5.3 \times 10^{- 3})$ $pH = 2.28$

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