1.

What will be the pH value of 0.05 M Ba(OH)_(2) solution

Answer»

12
13
1
12.96

Solution : `{:(Ba(OH)_(2),rarr,Ba^(+2)+,2OH^(-)),(,,0.05 M,2 xx 0.5 M):}`
`POH = LOG.(1)/([OH^(-)]) = log.(1)/(0.1) = 1`
`pH = pOH = 14 , pH + 1 = 14, pH = 14 -1 = 13`.


Discussion

No Comment Found

Related InterviewSolutions