What mass of sodium hydroxide should be used to yield a 100 mL solution with a pH of 12.28?

0.076 g

Explanation:

First obtain the hydroxide ion concentration from the pH of the solution;

Recall that pH + pOH = 14

Therefore;

pOH = 14-pH

pOH=14-12.28

pOH = 1.72

Since;

pOH = - log[OH^-]

[OH^-] = Antilog (-pOH)

[OH^-] = Antilog (-1.72)

[OH^-] = 0.019055 M

Note that

[Na^+]=[OH^-] = [NaOH] = 0.019055 M

But;

n = CV

Where;

n = number of moles given by mass / molar mass

C = concentration of solution

V = volume of solution

mass of NaOH = the unknown

Molar mass of NaOH = 40gmol-1

Let the mass of NaOH be m

Hence;

m/40 = 0.019055 * 100/1000

m/40 = 1.9055*10^-3

m = 1.9055*10^-3 * 40

m = 0.076 g