Calculate the [H+] ion of 0.008M Ca(OH)2 is?
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1 answer · Chemistry
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If we assume that the Ca(OH)2 is fully dissolved and dissociated ,
Then Ca(OH)2 ↔ Ca2+ + 2OH-
1mol Ca(OH)2 will produce 2 mol OH-
The solution is 8.8*10^-4M in Ca(OH)2 = (8.8*10^-4)*2 = 1.76*10^-3M on OH- ions
Equation:
[H+] [OH-] = 1*10^-14
[H+] = 1*10^-14 / [OH-]
[H+] = 1*10^-14 / 1.76*10^-3
[H+] = 5.68*10^-12M
Best Answer
If we assume that the Ca(OH)2 is fully dissolved and dissociated ,
Then Ca(OH)2 ↔ Ca2+ + 2OH-
1mol Ca(OH)2 will produce 2 mol OH-
The solution is 8.8*10^-4M in Ca(OH)2 = (8.8*10^-4)*2 = 1.76*10^-3M on OH- ions
Equation:
[H+] [OH-] = 1*10^-14
[H+] = 1*10^-14 / [OH-]
[H+] = 1*10^-14 / 1.76*10^-3
[H+] = 5.68*10^-12M
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But the answer is wrong
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