Question

How much NaOH (in grams) is needed to prepare 921 mL of solution with a pH of 10.405?

g NaOH

Answer 1

use:

pH = -log [H3O+]

10.4 = -log [H3O+]

[H3O+] = 3.936*10^-11 M

use:

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(3.936*10^-11)

[OH-] = 2.541*10^-4 M

So,

[NaOH] = 2.541*10^-4 M

volume , V = 921 mL

= 0.921 L

number of mol,

n = Molarity * Volume

= 0.0003*0.921

= 2.34*10^-4 mol

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH,

m = number of mol * molar mass

= 2.34*10^-4 mol * 39.998 g/mol

= 9.36*10^-3 g

Answer: 9.36*10^-3 g