Use the Henderson- Hasselbalch equation to calculate the pH of each solution.
a. a solution that is 0.145 M in propanoic acid and 0.115 M in potassium propanoate
b. a solution that contains 0.785% C5H5N by mass and 0.985% C5H5NHCI by mass
c. a solution that contains 15.0 g of HF and 25.0 g of NaF in 125 mL of solution
a)
pH = pKa + log{[salt]/[acid]}
pKa of propanoic acid = 4.87
[salt] = 0.115 M
[acid] = 0.145 M
=> pH = 4.87 + log(0.115/0.145)
=> pH = 4.77
b)
pOH = pKb + log{(moles of salt)/(moles of base)}
moles of base(C5H5N) = 0.785 / 79 = 0.00994 moles
moles of salt (C5H5NHCl) = 0.985 / 115.5 = 0.00853 moles
pKb of base ( C5H5N) = 8.77
=> pOH = 8.77 + log(0.00853/0.00994)
=> pOH = 8.70
But pH = 14 - pOH = 14 - 8.70
=> pH = 5.30
c)
pH = pKa + log{(moles of salt)/(moles of acid)}
moles of acid (HF) = 15.0 / 20 = 0.75 moles
moles of salt (NaF) = 25.0 / 42 = 0.595 moles
pKa of HF = 3.17
=> pH = 3.17 + log(0.595/0.75)
=> pH = 3.07
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