Answer
0.60 mL of this 15.0 % NaOH solution is needed to produce 5.00 L of an NaOH solution with pH = 10.8
Work Step by Step
1. Find the hydroxide concentration:
$$[H_3O^+] = 10^{-pH} = 10^{-10.8} = 2.0 \times 10^{-11} \space M$$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 2.0 \times 10^{-11} } = 5.0 \times 10^{-4} \space M$$
2. Since NaOH is a strong base: $ [NaOH] = [OH^-] = 5.0 \times 10^{-4} \space M$
3. Calculate the molar mass for NaOH:
$ NaOH $ : ( 22.99 $\times$ 1 )+ ( 1.008 $\times$ 1 )+ ( 16.00 $\times$ 1 )= 40.00 g/mol
4. Find the amount of moles necessary for 5.00 L of solution:
$$5.00 \space L \times \frac{5.0 \times 10^{-4} \space mol}{1 \space L} = 2.5 \times 10^{-3} \space moles$$
5. Find the volume of the initial solution that has this amount of moles:
$$2.5 \times 10^{-3} \space mol \space NaOH \times \frac{40.00 \space g \space NaOH}{1 \space mol \space NaOH} \times \frac{100 \space g \space solution}{15.0 \space g \space NaOH} \times \frac{1 \space mL \space solution}{1.116 \space g \space solution} = 0.60 \space mL \space solution$$
