Calculate the concentration of an aqueous solution of naoh that has a ph of 11.50

NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.

Therefore [NaOH] = [OH⁻]

To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.

pH + pOH = 14

Since pH = 11.50

pOH = 14 - 11.50

pOH = 2.50

We can calculate [OH⁻] by knowing pOH

pOH = - log[OH⁻]

[OH⁻] = antilog (-pOH)

[OH⁻] = 3.2 x 10⁻³ M

therefore [NaOH] = 3.2 x 10⁻³ M

We know that;

pH + pOH = 14

Thus; pOH = 14-11.5

= 2.5

But; pOH = - log [OH-]

-log[OH} = 2.5

[OH] = 10 ^-2.5

= 3.162 x 10^-3 M

thus; [NaOH] = 3.162 x 10^-3 M