11.Assuming complete dissociation ,Calculate the PH of the following solutions
(i)0.004 M HCl
(ii)0.005 M NaOH
Answers
Answered by
0
Answer:
11.301
Explanation:
(a) 0.003 M HCl
[H
3
O
+
]=[HCl]=0.003M
pH=−log[H
+
]=−log(3.0×10
−3
)=2.523
(b) 0.005 M NaOH
[OH
−
]=[NaOH]=0.005M
[H
+
]=
[OH
−
]
K
w
=
0.005
10
−14
=2×10
−12
pH=−log[H
+
]=−log(2×10
−12
)=11.699
(c) 0.002M HBr
[H
+
]=[HBr]=0.002
pH=−log[H
+
]=−log0.002=2.699
(d)0.002M KOH
[OH
−
]=[KOH]=0.002M
[H
+
]=
[OH
−
]
K
w
=
0.002
10
−14
=5×10
−12
pH=−log[H
+
]=−log(5×10
−12
)=11.301
hope it helps you
Similar questions