Question

#1

Given that at 25.0 ∘C

Ka for HCN is 4.9×10−10 and

Kb for NH3 is 1.8×10−5,

calculate

Kb for CN− and

Ka for NH4+.

Enter the Kb value for CN− followed by the Ka value for NH4+, separated by a comma, using two significant figures.

#2 Find the pH of a 0.250 M solution of NaC2H3O2. (The Ka value of HC2H3O2 is 1.80×10⁻⁵).

Answer 1

1.Since we know that at standard conditions at 25^oC

K_a*K_b=K_w=10^-14

so if K_a for HCN=4.9 * 10^-10

thus K_b for CN^-=10^-14/4.9 * 10^-10=2*10^-5

Similarly, if K_b form NH₃=1.8*10^-5

So,K_a for NH₄⁺=10^-14/1.8*10^-5=5.56*10^-10

2. HC₂H₃O₂+H₂O-->H₃O⁺+C₂H₃O₂^-, K_a=1.8*10^-5

But when NaC₂H₃O₂ is added in water,it acts as base

C₂H₃O₂^-+H₂O-->HC₂H₃O₂+OH^-, kb=kw/ka=5.56*10^-10

inital 0.25 0 0

equlibrium 0.25-x x x

thus,

let x<<0.25

thus

x=1.18*10^-5 M

[OH^-]=x=1.18*10^-5 M

pOH=-log([OH-])=4.92

pH+pOH=14

thus,pH=14-4.92=9.08