You have a flask in your laboratory on which the label is smudged. From what you

Question

You have a flask in your laboratory on which the label is smudged. From what you are able to read, it is a solution of a monoprotic acid (well call it HX) at a concentration of 2E-3 M in a 0.05 M NaCl solution. You measure the pH of the solution and get a reading of 3.09. Using this information, do the following:

(a) Determine the pKa of the acid and the identity of the X- using the information in Table 8.2. Recall that a pH meter measures activity, not concentration of protons.

(b) Determine the pH of a 4E-3 M NaX solution in 0.05 M NaCl

Explanation / Answer

Solution :-

a)Determine the pKa of the acid and the identity of the X- using the information in Table 8.2. Recall that a pH meter measures activity, not concentration of protons.

Solution :- pH= 3.09

Using the given pH lets calculate the concentration of the H^+ ions

pH= -log [H+]

[H+] = antilog [-pH]

         = antilog[ 3.09]

        = 8.128*10^-4 M

So the concentration of the X^- is also same as H+

Now lets calculate the ka

Ka = [H+][X-]/[HX]

     = [8.128*10^-4][8.128*10^-4] / [2*10^-3 – 8.128*10^-4]

    = 5.56*10^-4

Therefore ka of acid = 5.56*10^-4

(b) Determine the pH of a 4E-3 M NaX solution in 0.05 M NaCl

Solution :- lets calculate the kb of the X^-

Kb = kw/ ka

     =1*10^-14 / 5.56*10^-4

    = 1.80*10^-11

Now using the kb lets calculate the concentration of the OH-

Kb = [x][x]/[4*10^-3]

1.80*10^-11 = x^2 /4*10^-3

1.8*10^-11 * 4*10^-3 = x^2

7.2*10^-14 =x^2

Taking square root of both sides we get

2.68*10^-8 = x

Therefore

pOH = - log [OH-]

pOH = -log [ 2.68*10^-8]

pOH = 6.57

pH = 14- pOH

pH= 14-6.57

pH= 7.43

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