Calculate the [h30+] in a 0.010M solution of Sr(OH)2

Answer 1

0.010M in Sr(OH)2.
OH^- is twice that.
(H3O^+)(OH^-) = Kw = 1E-14
Solve for (H3O^+)

Answer 2

5*10•14

Answer 3

because Sr(OH)2, do 0.010 x 2

ans = 0.020

pOH = -log(0.020)
pOH = 1.7
[finding pH because Sr(OH)2]
pH = 14 - 1.7
pH = 12.30

10^-12.30 = 5.0x10^-13

hope this helps