Answer to Question #147213 in General Chemistry for Aidan

Question

We have 500.0 mL of a buffer solution which is 0.645 M in NH3 and 0.301 M in NH4Cl. 500.0 mL of a 0.216 M HCl solution are added. The final volume is 1.0000 L. The value of Kb for NH3 is 1.8 x 10-5. What is the final pH of this solution?

Answer 1

The final volume is 1.0000 L

                 = 1000 ml

                  

When NH3 reacts with HCl then the following reaction will occur

  NH3 + HCl -----> NH4Cl

500.0 mL 0.645 M in NH3 reacts with 500.0 mL of a 0.216 M HCl then NH4Cl produce.

So, total concentration of NH4Cl in the solution, [NH4Cl] = amount of NH4Cl ( produce by the reaction of NH3 and HCl + NH4Cl present initially)

[NH4Cl] =(500ml×0.216 M + 500ml×0.301 M)/final volume

       = (500ml×0.216 M + 500ml×0.301 M)/1000 ml

       = 0.2585 M

And amount of NH3 remain will be,

[NH3] = (500 ml×0.645 M - 500ml×0.216 M)/The final volume

  =(500 ml×0.645 M - 500ml×0.216 M)/1000 ml

  = 0.2145 M 

From the Henderson-hasselbalch equation of pH of buffer,

    pOH = pKb + log([NH4Cl]/[NH3])

  Or, pOH = -logKb + log([NH4Cl]/[NH3])

     = -log(1.8x10^-5)+log(0.2585/0.2145)

     = 4.825

We know, pH+pOH = pKw

      Or, pH = (pKw – pOH)

      Or, pH = (14–4.825)

      Or, pH = 9.175 

       

So, pH of the buffer solution is = 9.175