In: Chemistry
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
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