alculate the pH of 0.279 M nitrous acid. Nltrous acid (HNO2 ) is a weak acid, so

Question

alculate the pH of 0.279 M nitrous acid. Nltrous acid (HNO2 ) is a weak acid, so be sure to look up it's K. Show all the steps, be sure to include the chemical equation for the interaction of nitrous acid and water, the equilibrium expression for the K, of nitrous acid based on the reaction, and set up and fill in the equilibrium ("ICE" box). Calculate the pH of 0.366 M methylamine. Methylamine (CH3NH2) is a weak base, so be sure to look up it's K Show all the steps, be sure to include the chemical equation for the interaction of methylamine and water, the equilibrium expression for the Kb of methylamine based on the reaction, and set up and fill in the equilibrium ("ICE" box). (Remember the "x in the ice box is (OH]) This problem involves nitric acid (HNO2), acetic acid (CH COOH), potassium acetate (KCH,COO), and potassium hydroxide (KOH). a. What type of acid is HNO3? [strong or weak ?] Calculate the pH of a solution of 0.454 M HNO3- What type of acid is CH&COOH;? [strong or weak ?] Calculate the pH of a solution of 0.454 M CH3COOH. b. c. Which has a lower pH, 0.454M HNO, or 0.454M CH,COOH? d. Which has a lower [H'L, 0.454M HNO, or 0.454M CH,COOH? e. Calculate the pH of a solution of a solution that is 0.454M CH,COOH and 0.454M KCH,coo. (Hint: set up the equilibrium ("ICE" box) for CH3COOH first, then instead of "O" for initial CHC0 on the product side you will have 0.454M from the KCH,coo. Then calculate "x as usual for the [H'J.) f. What type of base is KCH,COO? [weak or strong ?] Calculate the pH of a solution of 0.454 M KCH3COO. What type of base is KOH? [weak or strong ?] Calculate the pH of a solution of 0.454M KOH. g.

Explanation / Answer

(a)

HNO3 is a strong acid and it will dissociate completely.

HNO3     =     H+     +     NO3-

Now,

[HNO3] = 0.454 M

So, [H+] = 0.454 M

pH = - log [H+]

      = - log (0.454)

      = 0.34

(b)

CH3COOH is a weak acid. It will dissociate as

CH3COOH     =     H+     +     CH3COO-

Here

[CH3COOH] = 0.454 M

Ka of CH3COOH = 1.8 x 10-5

Ka = { [H+] [CH3COO-] } / [CH3COOH]

1.8x10-5 = (x) (x) / (0.454 –x)

1.8x10-5 = x2 / (0.454)

x2 = (1.8 x 10-5) (0.454)

x2 = 8.17 x 10-6

x = 2.86 x 10-3

So, [H+] = 2.86 x 10-3

pH = - log[H+] =

      = - log (2.86 x 10-3)

      = 2.54

(c)

0.454 M HNO3 has lower pH, since HNO3 is a strong acid.

(d)

For, 0.454 M HNO3, [H+] = 0.454 M

For, 0.454 M CH3COOH, [H+] = 2.86 x 10-3 M

CH3COOH has lower [H+]

(f)

KCH3COO is a weak base. It will dissociate as

CH3COO-     +     H2O     =     OH-     +     CH3COOH

Here

[CH3COO-] = 0.454 M

Ka of CH3COOH = 1.8 x 10-5

Kb = Kw / Ka

     = (1 x 10-14) / (1.8 x 10-5)

    = 5.56 x 10-10

Kb = { [OH-] [CH3COOH] } / [CH3COO-]

5.56 x10-10 = (x) (x) / (0.454 –x)

5.56 x10-10 = x2 / (0.454)

x2 = (5.56 x10-10) (0.454)

x2 = 2.52 x 10-10

x = 1.59 x 10-5

So, [OH-] = 1.59 x 10-5

pOH = - log[OH-]

      = - log (1.59 x 10-5)

      = 4.80

pH = 14 – pOH

     = 14 – 4.80

    = 9.20

(g)

KOH is a strong base and it will dissociate completely.

KOH     =     K+     +     OH-

Now,

[KOH] = 0.454 M

So, [OH-] = 0.454 M

pOH = - log [OH-]

        = - log (0.454)

      = 0.34

pH + pOH = 14

pH = 14 – pOH

     = 14 – 0.34

    = 13.66

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