How to calculate pH of HCl in water?

Calculate the pH after the addition of 4.00mL of 0.100M HCl?

• For each of the following three questions 25.0 mL of 0.100 M pyridine (Kb = 1.70 x 10−9) is titrated with 0.100 M HCl. For each question calculate the pH of the final solution after the indicated amount of HCl has been added. That is, treat each question as a separate titration. Pyridine is monobasic, i.e., each molecule of pyridine accepts only one proton from a molecule of acid. Remember that the final volume of the solution is the sum of the initial volume and the volume of HCl that is added to it. Calculate the pH after the addition of 4.00 mL of 0.100 M HCl. Calculate the pH after the addition of 25.0 mL of 0.100 M HCl. Calculate the pH after the addition of 28.0 mL of 0.100 M HCl.

• Answer:

  pyridine (C5H5N) is a weak base and HCl is a strong acid... they react as : C5H5N + HCl -------> C5H5NH(+)Cl(-) or C5H5N.HCl where C5H5NH(+)Cl(-) or C5H5N.HCl is known as pyridine hydrochloride i.e. salt of pyridine (weak base) and HCl (strong acid).. as we know molarity = number of moles of solute/volume of solution in litres so number of moles of solute = molarity X volume of solution in litres so number of moles of pyridine = 0.1 X 0.025 = 2.5 X 10^-3 ..(25 ml = 25/1000 Litre = 0.025 Litre) 1.Calculating pH after adding 4.00 mL of 0.100 M HCl ------- number of moles of HCl = 0.004 X 0.1 = 4 X 10^-4 number of moles of pyridine in solution is 2.5 X 10^-3 but number of moles of HCl in the solution is 4 X 10^-4 ...so as HCl is present in lesser amount so here HCl is the limiting reagent.. now 4 X 10^-4 moles of HCl will react with 4 X 10^-4 mole of pyridine to produce 4 X 10^-4 mole of pyridine hydrochloride ... so number of moles of pyridine left = 2.5 X 10^-3 - 4 X 10^-4 = 2.1 X 10^-3 number of moles of pyridine hydrochloride = 4 X 10^-4 since pyridine is a weak base and pyridine hydrochloride is a salt of weak base and strong acid ...so they will form a basic buffer solution..and pH is given as : pH = pKb + log [salt]/[base] where [salt] = number of moles of salt in solution [base] = number of moles of base in solution putting the values.. pH = -log (1.7 X 10^-9) + log [4 X 10^-4]/[2.1 X 10^-3] pH = 8.7696 + log 0.1905 pH = 8.7696 - 0.7201 = 8.0495 2.Calculating pH after adding 25.0 mL of 0.100 M HCl. number of moles of HCl = 0.025 X 0.1 = 2.5 X 10^-3 number of moles of pyridine = 2.5 X 10^-3 so 2.5 X 10^-3 moles of HCl will react with 2.5 X 10^-3 moles of pyridine to produce 2.5 X 10^-3 mole of pyridine hydrochloride ... so we cannot use buffer equation as above as now no pyridine is left ..all pyridine is converted into pyridine hydrochloride ... volume of solution = 25 + 25 = 50 ml or 50/1000 = 0.05 Litre so concentration (molarity) of pyridine hydrochloride = 2.5 X 10^-3/0.05 = 0.05 mole/Litre using hydrolisis equation for a salt of weak base and strong acid ... pH = 1/2 [pKw - pKb - log c ] pH = 1/2 [ 14 - 8.7696 - log 0.05 ] pH = 1/2 [ 14 - 8.7696 - 1.301 ] pH = 1/2 [ 3.9294 ] = 1.9647 3.Calculating pH after adding 28.0 mL of 0.100 M HCl. number of moles of HCl = 0.028 X 0.1 = 2.8 X 10^-3 number of moles of pyridine = 2.5 X 10^-3 in this case number of moles of HCl is greater than number of moles of pyridine...so the excess of H(+) from the HCl will suppress the hydrolysis of C5H5NH(+) and we can ignore the contribution of H(+) from C5H5NH(+) and pH calculation will be done by H(+) from HCl only... so as number of moles of HCl = 2.8 X 10^-3 so number of moles of H(+) = 2.8 X 10^-3 total volume = 28 + 25 = 53 ml = 53/1000 Litre = 0.053 Litre so concentration of H(+) = 2.8 X 10^-3/0.053 = 0.0528 so pH = -log [H+] = -log [0.0528] pH = 1.2774