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