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Chapter 13: Problem 171

Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7}-M\) solution of \(\mathrm{NaOH}\) in water.

Short Answer

Expert verified
The pH of a \(1.0 \times 10^{-7} \ \text{M}\) solution of NaOH in water is 7.

Step by step solution

01

Write the dissociation reaction of NaOH in water

Write the reaction of NaOH dissociating into its ions in water: NaOH(aq) 鉄� Na鈦�(aq) + OH鈦�(aq)
02

Write the equilibrium reaction for self-ionization of water

Write the self-ionization of water reaction where water reacts with itself to form hydronium ions (H鈧僌鈦�) and hydroxide ions (OH鈦�): 2H鈧侽(l) 鈬� H鈧僌鈦�(aq) + OH鈦�(aq)
03

Find the concentration of OH鈦�

Since NaOH completely dissociates in water, the concentration of OH鈦� is equal to the concentration of NaOH: [OH鈦籡 = \(1.0 \times 10^{-7} \ \text{M}\)
04

Find the concentration of H鈧僌鈦� ions using equilibrium constant (Kw)

The ion product of water, \(K_w\), is the product of the concentration of H鈧僌鈦� ions and OH鈦� ions in the solution. At 25掳C, \(K_w = 1.0 \times 10^{-14}\). \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\) Rearrange the equation to find the concentration of H鈧僌鈦� ions: \([\text{H}_3\text{O}^+] = \dfrac{K_w}{[\text{OH}^-]}\) Insert the given values: \([\text{H}_3\text{O}^+] = \dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-7}}\) Solve for H鈧僌鈦� concentration: \([\text{H}_3\text{O}^+] = 1.0 \times 10^{-7} \ \text{M}\)
05

Calculate the pH of the solution

Use the formula for pH, which is \(pH = -\log_{10} [\text{H}_3\text{O}^+]\), to calculate the pH of the solution: \(pH = -\log_{10} (1.0 \times 10^{-7})\) Solve for pH: \(pH = 7\) So, the pH of a \(1.0 \times 10^{-7} \ \text{M}\) NaOH solution in water is 7.

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