l. What is the [H30 ]and [OH for milk if its pH is 6.40? B. What is the pH and p

Question

l. What is the [H30 ]and [OH for milk if its pH is 6.40? B. What is the pH and poH of the strong base 0.0030 M Ba(OH2? 2. A. What is the pk, for lactic acid if the pH of a 1.00 x 103 M solution is 3.23? Lactic Acid HLac CH3CH(OH)COOH B. What is the ionization dissociation) in this solution? C. Is the assumption that the self-ionization of water produces a negligible amount of Hyo ion valid here? How can you tell? Calculate the [oH) present! Water self. ionization produces equal amounts of oH and H30.

Explanation / Answer

1) if pH = 6.4

Then we know that

pH = -log[H+]

6.4 = -log[H+]

Taking antilog on both side

antilog of (-6.4) = [H+]

[H+] = 3.98 X 10^-7

We also know that

[H+] [OH-] = 10^-14

Therefore [OH-] = 10^-14 / [H+] = 10^-14 / 3.98 X 10^-7

[OH-] = 2.51 X 10^-8

b) for strong base Ba(OH)2

Each mole of Ba(OH)2 will give two moles of OH-

So the concentration of [OH-] = 2 X [Ba(OH)2] = 2 X 0.003 M = 0.006 M

Now for base

pOH = -log[OH-]

pOH = - log [0.006] = 2.22

pH = 14- pOH = 14 - 2.22 = 11.78

2) The dissociation of lactic acid will be

                 HLa --> La- + H+

Initial       10^-3         0       0

Change     -x            +x     +x

Equilibrium   10^-3 -x    x     x

Ka = [H+] [ La-] / [LaH]

given pH =3.23 = -log[H+]

[H+] = 0.000589 =x = [La-]

Ka = x^2 / 10^-3 - x

we may ignore "x" in deomintor as lactic acid is a weak acid and x <<1

Ka = x^2 / 10^-3

Ka = ( 0.000589)^2 / 0.001

Ka = 0.000347 = 3.47 X 10^-4

pKa = -logKa = 3.46

B) % dissociation = x X 100 / initial concentration = 0.000589 X 100 / 0.001 = 58.9 %

C) As the pH of solution is not so very high so we may ignore the self ionization here.

The OH- present = 10^-14 / [H+] = 10^-14 / 0.000589 = 1697.79 X 10^-14 = 1.697 X 10^-11

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