pH = 13.62
pOH = 0.38
(OH^-) = 0.417 M
M = moles/L; therefore,
moles = M x L = 0.417 x 0.1 L = 0.0417
grams = moles x molar mass = 0.0417 x 40 = ?? grams NaOH. I don't see 0.4.
What is the mass of NaOH required to prepare 100.0mL of NaOH(aq) that
has a pH = 13.62 ?
A. 0.38g
B. 0.42g
C. 1.67g
D. 2.40x10^-14 g
OH- + H2O <-> H3O+ + O^-2
antilog-13.62 = 0.09976
#gNaOH=(0.1-0.09976)M x 40g/mol x 0.1L = 0.40g?
2 answers
Genius up above me is wayyyy off answer is 1.67 g morons like him screw people over when they are they are trying to get help.
0 answers