Calculate the pH after adding 10.00 mL of 0.15 M HCl to 50 mL of the...

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

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Expert Solution

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

queen_honey_blossom answered 1 year ago

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