Question

If you made up a solution of NaOH by adding 0.010 mole of solid NaOH to 1.0 liter of distilled water, what would be the concentration of the OH-(aq)?

Answer 1

0.01M or 1 x10 ^-2M

Step-by-step explanation:

Answer 2

The concentration of the OH ^-(aq) is 0.01 M or 1 x10^-2 M


calculation
first write the dissociation equation for NaOH

NaOH → Na^+ + OH^-

find the molarity of NaOH used

molarity = moles/volume in liters= 0.010 moles/ 1L = 0.01 M

by use of mole ratio between NaOH to OH^- which is 1:1 therefore the concentration of OH^- is also 0.01M or 1 x10 ^-2M