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

Respuesta :

Answer:

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