ph of 0.005M BaoH2 is
Answers
Answered by
1
Answer:
concentration of OH
−
in the solution = 0.005×2=0.01mole/l
pOH=−log[0.01]=2
pH=14−pOH=14−2=12
When 100 ml of this solution diluted to 1000 ml
Moles of OH
−
=0.01 mole
Concentration of OH
−
=
1000
0.01×100
=0.001
pOH=−log[10
−3
]=3
pH=14−3=11
Explanation:
Similar questions