– ] / [NH ] = . × – Before neutralization, [NH + ] = [OH – ] = x [NH ] = . – x . x / .
× – pH = –log ( . × – ) = . On addition of mL of .1M HCl solution (i.e., . mmol of HCl) to mL of .1M ammonia solution (i.e., mmol of NH ), .
mmol of ammonia molecules are neutralized. The resulting mL solution contains the remaining unneutralized . mmol of NH molecules and . mmol of NH + .
NH + HCl → NH + + Cl – . . At equilibrium . .
The resulting mL of solution contains . mmol of NH + ions (i.e., . M) and . mmol (i.e., .
M ) of uneutralised NH molecules. This NH exists in the following equilibrium: NH OH NH + + OH – .033M – y y y where, y = [OH – ] = [NH + ] The final mL solution after neutralisation already contains . m mol NH + ions (i.e. .033M), thus total concentration of NH + ions is given as: [NH + ] = .
+ y As y is small, [NH OH] . M and [NH + ] .033M. We know, K b = [NH + ][OH – ] / [NH OH] = y ( . )/( .
× – Hence, pH = . . . Hydrolysis of Salts and the pH of their Solutions Salts