At what
what ph the
will occur
the precipitate La(oh)3
when
0.010M La with Naoh the ksp of
La(oh)3
2x10-21 ?
Answers
Answered by
18
Explanation:
0.010 M la3 is treated with naoh
The solution to know what is the ph:
La (OH) 3 = La3+ + 3OH-
Ksp = [La3+] [OH-^3
2 X 10^-21 = (0.010 M) [OH-] ^ 3
[OH-] = (2 X 10 ^ -21 / 0.010) ^1/3
[OH-] = 2.63 x 10^-7 M
= 5.85 x 10^-7 M
This occurs at pOH = 6.23 and pH = 7.77
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