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Basiskennis chemie 2/Basen/Opgaven
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Basiskennis chemie 2
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Basen
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Van [OH
-
] naar pOH
Bewerken
Bereken van onderstaande oplossingen de pOH in twee cijfers achter de komma.
1.
[
OH
−
]
=
8
,
25
⋅
10
−
14
{\displaystyle {\ce {[OH^{-}] \ = \ 8,25.10^{-14}}}}
Antwoord
2.
[
OH
−
]
=
7
,
50
⋅
10
−
2
{\displaystyle {\ce {[OH^{-}] \ = \ 7,50.10^{-2}}}}
Antwoord
3.
[
OH
−
]
=
9
,
85
⋅
10
−
11
{\displaystyle {\ce {[OH^{-}] \ = \ 9,85.10^{-11}}}}
Antwoord
4.
[
OH
−
]
=
9
,
01
⋅
10
−
5
{\displaystyle {\ce {[OH^{-}] \ = \ 9,01.10^{-5}}}}
Antwoord
5.
[
OH
−
]
=
9
,
65
⋅
10
−
7
{\displaystyle {\ce {[OH^{-}] \ = \ 9,65.10^{-7}}}}
Antwoord
6.
[
OH
−
]
=
1
,
13
⋅
10
−
9
{\displaystyle {\ce {[OH^{-}] \ = \ 1,13.10^{-9}}}}
Antwoord
7.
[
OH
−
]
=
7
,
72
⋅
10
−
3
{\displaystyle {\ce {[OH^{-}] \ = \ 7,72.10^{-3}}}}
Antwoord
8.
[
OH
−
]
=
3
,
69
⋅
10
−
12
{\displaystyle {\ce {[OH^{-}] \ = \ 3,69.10^{-12}}}}
Antwoord
9.
[
OH
−
]
=
4
,
78
{\displaystyle {\ce {[OH^{-}] \ = \ 4,78}}}
Antwoord
10.
[
OH
−
]
=
6
,
73
⋅
10
−
14
{\displaystyle {\ce {[OH^{-}] \ = \ 6,73.10^{-14}}}}
Antwoord
11.
[
OH
−
]
=
1
,
92
{\displaystyle {\ce {[OH^{-}] \ = \ 1,92}}}
Antwoord
12.
[
OH
−
]
=
9
,
13
⋅
10
−
13
{\displaystyle {\ce {[OH^{-}] \ = \ 9,13.10^{-13}}}}
Antwoord
13.
[
OH
−
]
=
2
,
10
⋅
10
−
3
{\displaystyle {\ce {[OH^{-}] \ = \ 2,10.10^{-3}}}}
Antwoord
14.
[
OH
−
]
=
1
,
00
⋅
10
−
9
{\displaystyle {\ce {[OH^{-}] \ = \ 1,00.10^{-9}}}}
Antwoord
15.
[
OH
−
]
=
2
,
70
⋅
10
−
7
{\displaystyle {\ce {[OH^{-}] \ = \ 2,70.10^{-7}}}}
Antwoord
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