complicated chem 2 questions
Started by htime07
2. of the compounds below, a 0.1 M aqueous solution of ________will have highest pH
a. KCN, Ka of HCN = 4.0 x 10^-10
b. NH4NO3, Kb of NH3=1.8 x 10^-5
c. NaOAc, Ka of HOAc = 1.8 x 10^-5
d. NAClO, Ka of HClO= 3.2 x 10^-8
e. NaHS, Kb of HS- = 1.8 x 10^-7
a. KCN, Ka of HCN = 4.0 x 10^-10
b. NH4NO3, Kb of NH3=1.8 x 10^-5
c. NaOAc, Ka of HOAc = 1.8 x 10^-5
d. NAClO, Ka of HClO= 3.2 x 10^-8
e. NaHS, Kb of HS- = 1.8 x 10^-7
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I think it should be square root (7.5E-12 x 0.15) then take the -log of the answer.
Same as above but just for review: pka=-log(Ka)
Large Ka=strong acid, small pKa=strong acid
You have to remember that if its Kb, you are finding pOH, and should substract the value from 14.
The way to approach the salt problems is to look at the conjugate acids and bases of the respective ions that form in solution. Ill do them one by one.
a. For this salt K+ and CN- are the conjugates, K+ is the conjugate acid of a strong base KOH. CN- is the conjugate base of a weak acid HCN. CN- is basic so we will form a basic solution.
b. NH4+is the conjugate acid of the weak base NH3 and NO3- is the conjugate base of the strong acid HNO3. no the NH4+ is slightly acidic due to the fact that NH3 is a weak base. so this solution will be slightly acidic.
c. d. e. For the rest of these look at it the same way. Just know your strong acids and bases they are not many. so as you can see we have Na+ in all which is the weak conjugate acid of the strong base NaOH. Now lets examine all the conjugate bases of those acids. Because they are all weak acids the conjugate base will be slightly moderate so all of them will form a basic solution.
Answer should be b. i hope. I didnt care to look at the Ka or Kb for this. Although u can ballpark them . Just know that the underlying concept that they are testing here is not how fast u can compute ph from the give Ka or Kb values but that strong bases will have weak conjugate acids and strong acids will have weak conjugate bases. Hope that helps
First write out the reaction.
X- + H2O -> HX + OH-
Then set up an ICE table.
You are given [X-]
____X- + H2O -> HX + OH-
I __0.15________ 0_____0
C __-x _________ +x___+x
E _0.15-x _______ x ____x
Since you're now dealing with [OH-] you must use Kb.
Kb=[HX][OH-] / [X-] = x^2 / 0.15
assuming 0.15-x ~ 0.15
1x10^-14 / 7.5x10^-12 = Kb = 1.33x10^-3
solve for x which is equal to [OH-]
1.33x10^-3 = x^2 / 0.15
1.33x10^-3 * 0.15 = x^2
take square root
x=0.014 = [OH-]
pOH = 1.85
14-pOH = 12.15
