Calculate the pH of 7.2 × 10-8 M HCl. Report your answer to the hundredths place. Next, What fraction of the total H in this solution is from the HCl? Report your answer to the hundredths place.

Answer 1

You need two ICE charts for this; i.e., one for the strong acid HCl and one for H2O.
HCl is a strong acid; therefore, it dissociates completely (100%).

.......HCl --> H^+ + Cl^-
I.....7.2E-8....0......0
C...-7.2E-8...7.2E-8...7.2E-8
E......0......7.2E-8..7.2E-8

........H2O --> H^+ + OH^-
I..............7.2E-8...0
C...............+x.....x
E...........7.2E-8+x....x

(H^+)(OH^-) = Kw
(7.2E-8+x)(x) = 1E-14
Solve this quadratic for x and (H^+) = 7.2E-8+x. Then pH from that.

fraction from HCl = (H^+) from HCl/total (H^+)
Post your work if you get stuck.

Answer 2

Thank you very much for the reply.

I was able to solve for X (X = 7.03E-8), which I added to that of HCl (Total = 1.405E-7) and was able to obtain the pH (6.85, assume this answer is deemed correct).

Here is my problem: I took the (H^+) from HCL (X?) and divided it by the (Total).

=(7.03E-8/1.405E-7)

I got an answer of: .5004 which when multiplied by 100 = 50.04%

That answer is incorrect, but I'm not sure where I have messed up.

Answer 3

I apologize, I saw my mistake, and used HCL (7.2E-8)/(Total) and got: 51.245%

This answer is still wrong, so now I am thoroughly confused.

Answer 4

I obtained 7.028E-8 for x. Add to 7.2E-8 = 1.4228E-8 for total. I would round that to 1.42E-7 and pH = 6.8468 which to the hundredths place is 6.85.
H^+ from HCl is not x. x is the H^+ from the ionization of H2O.
fraction from HCl = 7.2E-8/1.42E-7 = 50.70%. That's what I would report. I don't know when your prof rounds but if s/he rounds at the end then fraction is
(7.2E-8/1.4228E-7)*100 = 50.60%. In any case, however, I don't think the hundredths place is allowed since 7.2 only has two s.f. (unless you typed 7.20 as 7.2 :-).