<span>Same answer, different setup. We know that the sum of the oxidation numbers is zero for a compound and the ionic charge for a polyatomic ion, and we know that sulfite ion is -2.
Create an algebraic equation by multiplying the subscripts times the oxidation number of a single element.
+x -6 = -2
+x -2
S O3
Solve for x
x = +4</span>
The computation for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
Kb = [HA} [OH-] / [A-] where [A-] represents the concentration of CN- (.068M)
Kb = Kw / Ka = 1 x10-14 / 4.9 x 10-10 = 2 x 10-5
Since this is a salt solution which could be considered to have formed from the neutralization of a strong base (NaOH) and a weak acid (HCN), the Na+ will have no effect on the pH of the solution while the CN- ion will undergo hydrolysis:
CN- + H2O --> HCN + OH-
Based on this equation, the quantities of HCN and OH- produced must be the same and therefore [HCN]=[OH-]. We will set this equal to x.
Plugging into the original equation yields:
2 x 10-5 = x2 / .068 M
Solving for x yields 1.2 x 10-3 whidh is equal to the [OH-]
The pOH then is equal to -log (1.2x10-3) = 2.9
The pH of the solution would be 14 - 2.9 = 11.1
