Answer to Question #277350 in General Chemistry for hsh

Question

1. Initial concentration of N2H4CO fertilizer, is 1.95 M and K = 1.5x 10^ -14

N2H4CO (aq) + H2O(l) ⇋ H2NCONH3 ^+ (aq) + OHˉ(aq)

2. Initial concentration of 4.25 M aniline, C6H5NH2 with Kb = 3.8 x 10^-10

C6H5NH2 (aq) + H2O(l) ⇋ C6H5NH3 ^+(aq) + OHˉ(aq) 

Answer 1

Answer a)

pKb of C6H5NH2 = -log(3.8 x 10^-10) = 9.42

pH = 4.11

pOH = 14 - 4.11 = 9.89

According to Henderson Hasselbalch equation

pOH = pKb + log[C6H5NH3+]/[C6H5NH2]

9.89 = 9.42 + log[C6H5NH3+]/(0.50)

10^0.47 = [C6H5NH3+]/(0.50)

[C6H5NH3+]= 1.48 M (Answer)

b) Moles of NaOH = 3.2 g/(40 g/mol) = 0.08 mol

Moles of C6H5NH2 = 0.50 M*1.0 L = 0.50 mol

Moles of C6H5NH3+ = 1.48 M*1.0 L = 1.48 mol

NaOH reacts with C6H5NH3+ to form C6H5NH2.

So, final moles of C6H5NH2 = 0.50 + 0.08 = 0.58 mol

Final moles of C6H5NH3+ = 1.48 - 0.08 = 1.40 mol

Using Henderson Hasselbalch equation:

pOH = 9.42 + log(1.40/0.58)

= 9.80

pH = 14 - 9.80 = 4.20