An acidic solution containing 0.010 m la3 is treated with naoh until la(oh)3 precipitates. use the solubility product for la(oh)
3 to find the concentration of oh when la3 first precipitates. at what ph does this occur?
1 answer:
Given:
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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Answer: 1.31M
Explanation:
V1 = 10mL
C1 = 8.5M
V2 = 65mL
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C2 = (10 x 8.5 ) /65
C2 = 1.31M
