## Chemistry: Molecular Approach (4th Edition)

1. Calculate the molar mass: $NaOH$ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol 2. Use the information as conversion factors to find the molarity of this solution: $$\frac{ 1.55 g \space NaOH }{100 \space g \space solution} \times \frac{1 \space mol \space NaOH }{ 40.00 \space g \space NaOH } \times \frac{ 1.01 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.391 \space M$$ 3. Find the pH of the solution. NaOH is a strong base: $[OH^-] = [NaOH] = 0.391 \space M$ $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.391 } = 2.56 \times 10^{-14} \space M$$ $$pH = -log[H_3O^+] = -log( 2.56 \times 10^{-14} ) = 13.592$$