A chemist boils off the water from 627 mL of a solution with a pH of 10.2. The only dissolved compound in the solution is NaOH. How much solid remains

4.03 x 10∧-3

Explanation:

The pH of the solution is 10.2, and we know:

pH + pOH = 14

pOH = 14 - pH

pOH = 14 - 10.2 = 3.8

pOH = - log[OH-]

3.8 = - log[OH-]

-3.8 = log[OH-]

[OH-] = 10∧-3.8 = 1.58 * 10∧-4

since NaOH (s) → Na∧ + (aq) + OH∧ - (aq), [NaOH] = 1.58*10-4.

627mL * 1L/1000mL * 1.58*10-4molesNaOH/1L * 40g NaOH/1mole NaOH

=4.03 * 10∧-3g NaOH