In: Chemistry
Calculate the pH after adding 10.00 mL of 0.15 M HCl to 50 mL of the ammonia/ammonium ion buffer solution with a ph of 4.82
The solution in which HCl is added is a buffer
The pH of buffer is given as
pH = pKa + log [salt] / [acid]
Or for basic buffer (like ammonia)
pH = pKa + log [base] / salt]
the moles of acid added = molarity X volume = 0.15 X 10 / 1000 = 0.0015 moles
pKb for Ammonia = 4.74
pKa = 9.26
[I beleive the pH should not be 4.82 as it is a basic buffer (alkaline), it must be the pOH = 4.82 therefore pH =14-4.82 = 9.18
pH = pKa + log [NH4Cl] / [NH3]
9.18 = 9.26 + log [NH4Cl] / [NH3]
[NH4Cl] =0.832 [NH3]
Now we have added Hcl to it, which will react with NH3 to form equivalent moles of NH4Cl and will decrease the concentration of NH3
Moles of HCl added = 0.15 X 10 / 1000 = 0.0015
[NH4Cl] = [NH4Cl + 0.0015] = 0.832 [NH3] + 0.0015
[NH3] = [NH3 - 0.0015]
new pH = pKa + log [0.832 [NH3] + 0.0015 ] / [NH3 - 0.0015]
So if we know the initial concentration of buffer solution we can calculate the pH of new solution
The pH will be > previous pH