Question

A 24.3 mL sample of 0.294 M ammonia, NH3, is titrated with 0.310 M hydrochloric acid. At the titration midpoint, the pH is Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

Answer :

Reaction is

NH_{​​​​​3} + HCl ---> NH_{​​​​​4} Cl

In the titration as we start putting acid into the beaker containing ammonia , NH_{​​​​​4} Cl start forming .

NH_{​​​​​3} and NH_{​​​​​4}​​​​Cl we make a basic buffer solution.

For basic buffer solution , Henderson Hesselbatch equation is applied to calculate pH of solution .

pOH = pK_{​​​​​b} + log ( [NH_{​​​​​4} Cl ] / [NH_{​​​​​3} ] )

At the titration midpoint , where exactly half of the ammonia gets neutralised with HCl to form NH_{​​​​​4} Cl . At the half point , concentration of NH_{​​​​​3} and NH_{​​​​​4} Cl will be equal.

So, from Henderson Hesselbatch equation

at half point

pOH = pK_{​​​​​b}

K_{​​​​​​b} of NH_{​​​​​3} ( from Google ) = 1.8 * 10^-5

pK_{​​​​​b} = - log(K_{​​​​​​b} ) = - log ( 1.8 * 10^-5 ) = 4.74

So, pOH = 4.74

pH = 14 - pOH = 14 - 4.74 = 9.26

So, pH of solution at midpoint = 9.26