C[NH4Cl]= 0.1 mol L-1
C[H] = 4/1 x C[NH4Cl]
C[H]= 0.4 mol L-1
PH= -log(0.4) =0.4
C[H] = 4/1 x C[NH4Cl]
C[H]= 0.4 mol L-1
PH= -log(0.4) =0.4
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i'm afraid that's not correct
after dissolution you have NH4+ and Cl-. only NH4+ is relevant
the concentration is c[NH4+] = c[NH4Cl] = 0.1mol/L
NH4+ is the acid and react following this reaction:
NH4+ + H2O -> NH3 + H3O+ (or H+ & H2O if you want)
unfortunately NH4+ is a weak acid (pKa = 9.25), so it don't dissociate totally c[NH4+] =/= c[H+] and you have to determine the dissociation-constant α first
α = √(Ks / c[NH4+]) this is just an approximation, the correct equation is a total pain
pKa[NH4+] = 9.25 => Ka = 10^-9.25
α = √(10^-9.25 / 0.1) = 7.5^-5 (which means that only 0.0075% of the acid is dissociated)
pH = -log(α * c[NH4+]) = -log(7.5^-5 * 0.1) = 5.125
after dissolution you have NH4+ and Cl-. only NH4+ is relevant
the concentration is c[NH4+] = c[NH4Cl] = 0.1mol/L
NH4+ is the acid and react following this reaction:
NH4+ + H2O -> NH3 + H3O+ (or H+ & H2O if you want)
unfortunately NH4+ is a weak acid (pKa = 9.25), so it don't dissociate totally c[NH4+] =/= c[H+] and you have to determine the dissociation-constant α first
α = √(Ks / c[NH4+]) this is just an approximation, the correct equation is a total pain
pKa[NH4+] = 9.25 => Ka = 10^-9.25
α = √(10^-9.25 / 0.1) = 7.5^-5 (which means that only 0.0075% of the acid is dissociated)
pH = -log(α * c[NH4+]) = -log(7.5^-5 * 0.1) = 5.125