For (a) so you know HNO3 is a strong acid meaning that it ionizes 100% so the (H+) from HNO3 will be 0.1 M. You know HCN is a weak acid with a Ka of 6.2E-10. That tells you that the H+ contributed by the HCN is VERY small from HCN ==> H^+ + CN^-. Then to make it even smaller, the high concentration of H^+ from the HNO3 shifts that HCN equilibrium to the LEFT which makes the H^+ contribution from HCN EVEN SMALLER. So you ignore the H^+ from the HCN. So HNO3 contribution is 0.1 M and that gives you a pH of 1.00. BTW, what is that M by Ka? Two points. First, Ka has no units. Some may argue with that but technically that is correct since Ka is derived from activity coefficients and those have no units. Second, even if you scrape units out of a Ka (using formula units) I don't know of any that has units of M = molar. For b you have a solution of HNO2 and HCN. This is a weak acid (HNO2) and another weak acid (HCN). HNO2 is the stronger of the two (from the Ka values) PLUS it is five times stronger in molarity so again I would ignore the HCN.
...................HNO2 ==> H^+ + NO2^-
I....................5...............0..........0
C...................-x..............x...........x
E...................5-x..............x...........x
Ka = 4E-4 = (H^+)(NO2^-)/(HNO2)
4E-4 = (x)(x)/(5-x) Solve for x = about 0.044 (My calculator isn't working but that's about right and pH is about 1.35.
(c) is an old standby chemistry trick. Many students will give a pH of 10 BUT remember that is in H2O where the (H^+) = (OH^-) [without the HCl] so that becomes
...................HOH ==> H^+ + OH^-
I................................1E-7.........1E-7
add............................1E-10..........0
C.................................+1E-10......?
E.................................1E-7
In other words the H^+ from the HCl contributes so little to the 1E-7 already there that the final is essentially 1E-7 or pH = 7.00.
In mixtures like this you must think your way through them; i.e., ignore the one contributing so little. Remember, too, that you may have a mixture of a weak acid/salt or weak base/salt and these you tackle with the Henderson-Hasselbalch equation. Hope this helps.
Calculate the pH of (a) a solution that contains 0.10 M HNO3(aq) (Ka ~20 M) and 0.10 M HCN(aq) (Ka = 6.2 × 10–10 M), (b) a solution that contains 1.00 M HCN(aq) (Ka = 6.2 × 10–10 M) and 5.00 M HNO2(aq) (Ka = 4.0 × 10–4 M), and (c) 1.0 × 10–10 M HCl (aq). (Answer: (a) 1.00, (b) 1.35, (c) 7.00)
I know the answers to these problems but not really the process. I know how to individually solve for 0.10M HNO3 and 0.10M HCN but not together.
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