The P^(H) of 0.005 M Ba (OH)_(2) is

1.	The P^(H) of 0.005 M Ba (OH)_(2) is
Answer» `2.301` `11.699` `12` `7` SOLUTION :`0. 00 5 M " Ba "(OH)_2 , N = 5 xx 10 ^(-3)xx 2 = 10 ^(-2) ` ` [OH^(-) ] = 10 ^(-2)RARR [H^(+) ] = 10 ^(-12)rArr pH = 12`

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