Question

If 5.2 g of butanoic acid, C4H8O2, is dissolved in enough water to make 1.0 L of solution, what is the resulting pH?

Assuming complete dissociation, what is the pH of a 4.41 mg/L Ba(OH)2 solution?

The Kb for an amine is 9.747 × 10-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.825? (Assume that all OH– came from the reaction of B with H2O.)

Answer 1

Mol. wt of Butanoic acid = 88.11 g·mol⁻¹
no. mole of Butanoic acid = mass/mol.wt
                          = 5.2 g/ 88.11 g·mol⁻¹
                          = 0.059 mol
Since, we have 1 L solution; the molarity of butanoic acid = 0.059 mol/L = 0.059 M
                      
                      
Butanoic Acid Ka Value= 1.5x10^-5

           C4H7COOH + H2O → H3O+1 + C4H7COO-1
Initial       0.059               0         0
change       -x                           +x          +x
Equilibrim   0.059-x            x            x

Ka = [C4H7COO-][H3O+]/[C4H7COOH]
1.5x10^-5 = x2/(0.059 -x) (since, x is very small; we can take 0.059-x ~ 0.059 )
x² = (1.5x10^-5)(0.059)
x² = 8.85 x 10^-7
x = 9.4 x 10^-4

Therefore [H3O+] =9.4 x 10^-4 M
pH = -log[H3O⁺] = 3.03

Assuming complete dissociation, what is the pH of a 4.41 mg/L Ba(OH)₂ solution?

Ba(OH)2 <==> Ba2+ + 2OH-

no. moles of Ba(OH)2 = (4.41 x 10^-3 g)/     (171.34 g/mol )
                   =2.574 x 10^-5 moles/L
                  
Therefore, no. moles of OH- = 2 x 2.574 x 10^-5 moles
                           = 5.148 x 10^-5 moles.
Since, the solution volume is 1L;
the molarity of [OH-] = 5.148 x 10^-5 M

pOH = -log[OH-] = -log[5.148 x 10^-5 M]
    = 4.29
  
Therefore, pH + pOH = 14
pH = 14- pOH
   = 14 - 4.29
   = 9.71

Q3> The Kb for an amine is 9.747 × 10^-5. What percentage of the amine is protonated if the pH of a solution of the amine is 9.825?
Henderson-Hasselbalch equation:
    pOH = pKb + log[BH⁺]/[B]
  
   pH + pOH = pKw
   pOH = 14 - 9.825 = 4.175
  
   pKb = -log(Kb)
       = -log(9.747 × 10^-5)
       = 4.01
Henderson-Hasselbalch equation:
    pOH = pKb + log[BH⁺]/[B]  

   4.175 = 4.01 + log[BH⁺]/[B]
   log[BH⁺]/[B] = 0.165
   [BH⁺]/[B] = 1.462
  
   [BH⁺]= [B] x 1.462   
  
   % BH⁺ protonated = ([BH⁺]/([B] + [BH⁺]))*100%
                      = (1.462[B]/([B] + 1.462[B])*100%
                       = (1.462/2.462)*100%
                       = 59.4% protonated