Question

11.Assuming complete dissociation ,Calculate the PH of the following solutions

(i)0.004 M HCl

(ii)0.005 M NaOH​

Answer 1

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