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)?
You have 0.01 mol NaOH/L and since there is 1 OH in 1 NaOH, the concn of OH is 0.01 mol/L = 0.01 M.
To calculate the concentration of OH-(aq) in the solution, we need to determine the number of moles of OH- ions present in the 1.0 liter solution. We can use the balanced chemical equation for the dissociation of NaOH in water:
NaOH (s) -> Na+ (aq) + OH- (aq)
According to the equation, one mole of NaOH produces one mole of OH-. Therefore, since we added 0.010 mole of NaOH to the solution, we also have 0.010 mole of OH- ions.
The concentration of OH-(aq) is defined as the moles of OH- ions divided by the volume of the solution. In this case, the volume of the solution is 1.0 liter.
Concentration of OH-(aq) = Moles of OH- ions / Volume of solution
Concentration of OH-(aq) = 0.010 mole / 1.0 liter
Therefore, the concentration of OH-(aq) in the solution would be 0.010 M (moles per liter).