can someone pls explain

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rtvj

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the ph of a soln having a 10 -5 M concentration of OH- is
1.5
2.7
3.9

the answer given is 9..can someone explain how the answer was calculated.

thanx
rtvj
 
rtvj said:
the ph of a soln having a 10 -5 M concentration of OH- is
1.5
2.7
3.9

the answer given is 9..can someone explain how the answer was calculated.

thanx
rtvj

PH + POH = Kw= 14
hence POH=14 - PH
FOR THE ABOVE QUESTION THE PH = 5
HENCE POH = 14 - 5 = 9
 
hi rtvj,
kw is constant for water which is always 10-14....so when u take a neg log it becomes 14.....and one value will b given.
by using,
kw = pH + pOH.........

cheers......
bye.
 
thanks both of u🙂

ssaug said:
hi rtvj,
kw is constant for water which is always 10-14....so when u take a neg log it becomes 14.....and one value will b given.
by using,
kw = pH + pOH.........

cheers......
bye.
 
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