Question

You need a strong acid for an experiment, and you are given a bottle that you are told contains a strong acid at 15.0 M concentration. (The name of the acid on the label was rubbed off, so you are not sure of the identity of the acid.) You make a 0.1 M solution and you want to know if you can assume that the [H+] will be 0.1 M since the acid is a strong acid and thus should (so you have been told) completely dissociate. You see that the label on the bottle lists a pKa for the acid of -6.3. A pKa? Why does a strong acid have a pKa? You decide that you better calculate the relative error that will result when you make the assumption that a 0.1 M solution of the strong acid will result in a 0.1 M [H+] versus the concentration that you obtain when you properly treat the dissociation using the listed pKa value. Normally, the relative error is expressed as follows: Er = ((xi – xt)/xt) × 100; where xi is the value under the assumption that complete dissociation occurs and xt is the true or actual value obtained when the pKa of the acid is taken into account. Now, the units for an answer that is expressed as a percent could equivalently be stated as “parts per hundred”. Because the value of ((xi – xt)/xt) is likely to be a small number, I want you to multiply this value by 109rather than 102 (as shown in the above equation). Thus, your answer will be in terms of parts per billion (ppb), and you should report your answer in these terms to the nearest ones (e.g., enter “11” if your answer is 11 ppb). For this problem, you can ignore the contribution of H+ ions due to the autoproteolysis of water. You will need to do this problem with the aid of a spreadsheet, and be sure to adjust the format of the cells to retain several figures beyond the decimal point.

Answer 1

1. pKa of stong acids means that hydronium ion [H+] concentration in aqeous solution at standard pressure and temp is equal to the strong acid introduced to the solution.

2.pH = -log(0.1) = 1 i.e [H+] = 0.1

3. pKa = -log (Ka) = pH for strong acids as the concentration of [H+] = [A-]

so, pH = pKa = -6.3 = -log(H+)

[H+] = 1995262 at 15 M concentration

At 0.1 M, it would be 1995262/150 = 1330

4. Er = ((xi – xt)/xt) × 100

where,

xi is the value under the assumption that complete dissociation occurs and

xt is the true or actual value obtained when the pKa of the acid is taken into account.

= [(0.1-1330.41)/1330.41] X 1000000000 = - 999924835 ppb

[acid]	0.1M
pH value	1
pKa	-6.3
error	- 999924835 ppb