Question

In: Chemistry

What is the pH of a 1.00 L buffer solution that is 0.80 mol L-1 NH3...

What is the pH of a 1.00 L buffer solution that is 0.80 mol L-1 NH3 and 1.00 mol L-1 NH4Cl after the addition of 0.070 mol of NaOH(s)? Kb(NH3) = 1.76 ? 10?5

Solutions

Expert Solution

This is the basic buffer constituting NH3 & NH4Cl

According to Henderson's Equation ,

       pOH = pKb + log([salt]/[base])

    14-pH = pKb + log([salt]/[base])

         pH = 14 - pKb - log([salt]/[base])

              = 14 -(-logKb) - log([NH4Cl]/[NH3])

             = 14 - (-log(1.76 x 10-5)- log(1.00/0.80)

            = 14 - 4.75 - 0.097

            = 9.15

When 0.070 mol of NaOH is added :

             NH4Cl + NaOH NH3 + NaCl + H2O

[ NH4Cl ] = 1.00 - 0.070 = 0.93 M

[NH3] = 0.80 + 0.070 = 0.87 M

According to Henderson's Equation ,

         pH = 14 - pKb - log([salt]/[base])

              = 14 -(-logKb) - log([NH4Cl]/[NH3])

             = 14 - (-log(1.76 x 10-5)- log(0.93/0.87)

            = 14 - 4.75 - 0.029

            = 9.22

Therefore the pH of the solution after addition of NaOH is 9.22


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