6) Find the pH of the following solutions:

a) 0.00010 M HClO4

b) 2.35 x 10-3 M HI

c) 4.678 x 10-5 M HNO3

d) 0.0000877 M HCl

e) 0.00020 M HF

f) 0.0000061 M H3O+

g) 4.3 x 10-3 M NaOH (This is a base! Remember you will be finding pOH first!)

h) 6.7 x 10-9 M KOH (This is a base! Remember you will be finding pOH first!)

i) 0.00044 M HCl

j) C

pH is the negative logarithm of hydrogen (oxonium) ion concentration of aqueous solutions.

Explanation:

Recall that pH is defined as the negative logarithm of hydrogen ion concentration. We are going to apply that definition here.

a) 0.00010 M HClO4

pH= -log[0.00010]

pH= 4

b) 2.35 x 10-3 M HI

pH= -log[2.35 x 10-3 ]

pH= 2.63

c) 4.678 x 10-5 M HNO3

pH= -log[4.678 x 10-5]

pH= 4.3

d) 0.0000877 M HCl

pH=-log[0.0000877]

pH= 4.1

e) 0.00020 M HF

pH= -log[0.00020]

pH= 3.7

f) 0.0000061 M H3O+

pH= -log[0.0000061]

pH= 5.2

g) 4.3 x 10-3 M NaOH

pOH= -log[OH^-]

pOH= -log[ 4.3 x 10-3 ]

pOH= 2.4

But

pH + pOH =14

Therefore

pH= 14-pOH

pH= 14 -2.4

pH= 11.6

h) 6.7×10^-9 M KOH

pOH= -log[OH^-]

pOH= -log[ 6.7 x 10-9 ]

pOH= 8.2

But

pH + pOH =14

Therefore

pH= 14-pOH

pH= 14 -8.2

pH=5.8

I) 0.00044M HCl

pH= -log[0.00044]

pH= 3.4

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output?

explanation: