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ChemistryChemistry
24 November, 01:34

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

+2
Answers (2)
  1. 24 November, 01:50
    0
    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
  2. 24 November, 01:58
    0
    Answer: A. 4 B. 2.6 C. 4.3 D. 4.1 E. 3.7 F. 5.8 G. 11.6 H. 5.8 I. 3.4

    Explanation:

    A. 0.00010 M HClO4

    pH = - log [1.0 x 10^-4]

    = 4 - log 1 = 4

    B. 2.5 x 10^-3 M HI

    pH = - log [2.5 x 10^-3]

    = 3 - log 2.5

    = 3 - 0.398 = 2.6

    C. 4.678 x 10^M HNO3

    pH = - log [4.678 x 10^-5]

    = 5 - 0.7 = 4.3

    D. 0.0000877M HCL

    pH = - log [8.77 x 10^-5]

    = 5 - 0.943 = 4.057 = 4.1

    E. 0.00020M HF

    pH = - log [2.0 x10^-4]

    = 4 - 0.301 = 3.7

    F. 0.0000061M H3O+

    pH = - log [ 6.1 x 10^-6]

    = 6 - 0.785 = 5.2

    G. 4.3 x 10^-3 M NaOH

    Finding pOH first, we have

    pOH = - log [4.3 x 10-3]

    = 3 - 0.633 = 2.4

    pH + pOH = 14

    pH = 14 - 2.4 = 11.6

    H. 6.7 x 10^-9 M KOH

    Finding pOH, we have,

    pOH = - log [6.7 x 10^-9]

    = 9 - 0.826 = 8.17 = 8.2

    pH + pOH = 14

    pH = 14 - 8.2 = 5.8

    I. 0.00044 M HCl

    pH = - log [ 4.4 x 10^-4]

    = 4 - 0.64 = 3.36 = 3.4
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Home » Chemistry » 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.
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