## General Chemistry: Principles and Modern Applications (10th Edition)

$$pH = 13.74$$
1. Calculate the molar mass: $KOH$ : ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )+ ( 39.10 $\times$ 1 )= 56.11 g/mol 2. Use the information as conversion factors to find the molarity of this solution: $$\frac{ 3.00 g \space KOH }{100 \space g \space solution} \times \frac{1 \space mol \space KOH }{ 56.11 \space g \space KOH } \times \frac{ 1.0242 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.548 \space M$$ Since $KOH$ is a strong base: $[KOH] = [OH^-]$ 3. Calculate the pH: $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.548 } = 1.8 \times 10^{-14} \space M$$ $$pH = -log[H_3O^+] = -log( 1.8 \times 10^{-14} ) = 13.74$$