18)
Given:
pH = 10
use:
pH = -log [H+]
10 = -log [H+]
[H+] = 1*10^-10 M
use:
[OH-] = Kw/[H+]
Kw is dissociation constant of water whose value is 1.0*10^-14 at
25 oC
[OH-] = (1.0*10^-14)/[H+]
[OH-] = (1.0*10^-14)/(1*10^-10)
[OH-] = 1*10^-4 M
So,
[KOH] = 1*10^-4 M
Consider 1 L of solution
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 1*10^-4*1
= 1*10^-4 mol
Molar mass of KOH,
MM = 1*MM(K) + 1*MM(O) + 1*MM(H)
= 1*39.1 + 1*16.0 + 1*1.008
= 56.108 g/mol
use:
mass of KOH,
m = number of mol * molar mass
= 1*10^-4 mol * 56.11 g/mol
= 5.611*10^-3 g
Answer: option 1
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