Question

Use the Henderson–Hasselbalch equation to calculate the pH of each solution:

Part A a solution that is 0.155 M in propanoic acid and 0.120 M in potassium propanoate Express your answer using two decimal places. Part B a solution that contains 0.755% C5H5N by mass and 0.810% C5H5NHCl by mass Part C a solution that is 16.5 g of HF and 22.0 g of NaF in 125 mL of solution

Answer 1

1.pH= pKa + log [ conjugate base/acid]

pKa of propanoic acid = 4.87

pH= 4.87 + log (0.120/0.155) =4.76

2

Kb for CH3NH2= 1.7*10^-9

pKb= 8.77

C5H5NHCl + H2O ==⇒ C5H5N + H3O+

Basis : 1 Liter

0.755 %C5H5N = 0.00785(1000) = 7.55 g

0.810 % C5H5NHCl = 0.00810(1000) = 8.10 g

Moles of C5H5N= mass/ moles =7.55/79 = 0.09557, moles of C5NHCL =8.1/116=0.06983

pH = 14 - (pKb + log (salt/base)) =14- (8.77+ log (0.06983/0.09557) =5.36

3. moles of HF=mass/ molar mass = 16.5/20 =0.825 concentration =0.825*1000/125 Moles/L =6.6 M

moles of NaF= 22/42=0.524 concentration = 0.524/0.125 =4.2 M

pH= pKa + log [salt/acid], pka of HF= 3.14

pH= 3.14+ log (4.2/6.6)=2.94