if 0.050 mol of hcl is added to 1.0 l of buffer solution containing 0.125 M...

if 0.050 mol of hcl is added to 1.0 l of buffer solution containing 0.125 M potassium acetate and 0.10 M acetic acid what is the change in PH of the solution

Expert Solution

Potassium acetate reacts with HCl to produce acetic acid as per the reaction

CH3COOK + HCl -------->CH3COOH+ KCl

moles of potassium acetate (CH3COOK) in 1 L of 0.125M= Molarity* Volume(L)=0.125moles

moles of HCl = 0.050 moles, moles of acetic aicd in the buffer= 0.1*1=0.1 moles

HCl will be the limiting reactant,all the HCl gets consumed moles of acetic acid formed from the reaction =0.050

total of acetic acid =0.1+0.050=0.15, moles of potassium acetate remaining = 0.125-0.05= 0.075 moles

pKa of acetic acid= 4.76

pH= pKa+log [ potassium acetate]/[Acetic acid]

moles ratio of potassium acetate to acetic acid is equal to concentration ratio of potassium acetate to acetic acid

= 4.76+log(0.075/0.15)= 4.46

queen_honey_blossom answered 2 years ago

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