stupidatmath7829 stupidatmath7829
  • 03-12-2017
  • Chemistry
contestada

Estimate the solubility of m(oh)2 in a solution buffered at ph = 7.0, 10.0, and 14.0.

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Kalahira
Kalahira Kalahira
  • 15-12-2017
When M(OH)2 dissolves we have M(OH)2 which produces M2+ and 2OHâ’ pH + pOH=14 At ph =7; we have 7+pOH=14 pOH=14â’7 = 7 Then [OHâ’]=10^(â’pOH) [OH-] = 10^(-7) = 1* 10^(-7) At ph = 10. We have, pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4) Finally ph = 14. We have, pOH = 0 And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)... So [OH-] = 1
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