Answer
The pH of this KOH solution in 13.842
Work Step by Step
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:
$$\frac{ 3.85 g \space KOH }{100 \space g \space solution} \times \frac{1 \space mol \space KOH }{ 56.11 \space g \space KOH } \times \frac{ 1.01 \space g \space solution}{1 \space mL \space solution} \times \frac{1000 \space mL}{1 \space L} = 0.693 \space M$$
3. Calculate the pH of a KOH solution with this concentration.
Since KOH is a strong base: $[OH^-] = [KOH] = 0.693 \space M$
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.693 } = 1.44 \times 10^{-14} \space M$$
$$pH = -log[H_3O^+] = -log( 1.44 \times 10^{-14} ) = 13.842 $$
