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saul85 [17]
3 years ago
14

What are the pH of these solutions?

Chemistry
1 answer:
Tju [1.3M]3 years ago
3 0

Answer:

The answer to your question is below

Explanation:

a)    HCl  0.01 M

      pH = -log [0.01]

      pH = - (-2)

     pH = 2

b)    HCl = 0.001 M

      pH = -log[0.001]

      pH = -(-3)

      pH = 3

c)    HCl = 0.00001 M

       pH = -log[0.00001]

       pH = - (-5)

      pH = 5

d) Distilled water

      pH = 7.0

e) NaOH = 0.00001 M

       pOH = -log [0.00001]

       pOH = -(-5)

       pH = 14 - 5

       pH = 9

f)  NaOH = 0.001 M

      pOH =- log [0.001]

      pOH = 3

      pH = 14 - 3

      pH = 11

g)   NaOH = 0.1 M

       pOH = -log[0.1]

       pOH = 1

       pH = 14 - 1

       pH = 13

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Explanation:

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                  K.E_{2} = \frac{hc}{\lambda} - \phi  

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Now, divide equation (2) by 2. Therefore, it will become

       {6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}

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Now, subtract equation (3) from equation (1), we get the following.

                 1.6 \times 10^{-19} = \frac{\phi}{2}

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Thus, we can conclude that work function of the metal is 2 eV.

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