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
