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 = pK_b + log([salt]/[base])

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

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

= 14 -(-logK_b) - 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 - pK_b - log([salt]/[base])

= 14 -(-logK_b) - 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

queen_honey_blossom answered 1 year ago

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