Bonsoir,
c) On sait que:
pH=pKa+log([base]/[acide])
Comme on a: [base]=10[acide] donc:
pH=pka+log(10[acide]/[acide])
pH=pKa+log(10)
pKa=pH-log(10)
pKa=11.7-log(10)
pKa=11.7-1
pKa=10.7
d) On reprend la formule:
pH=pKa+log([base]/[acide])
pH=pKa+log((n(base)/(V(solution)[acide]))
log((n(base)/(V(solution)[acide]))=pH-pKa
n(base)/(V(solution)[acide])=10^(pH-pKa)
[acide]=n(base)/[V(solution)×10^(pH-pKa)]
[acide]=0.001/(0.5×10^(9.7-10.7))
[acide]=0.02 molL⁻¹
[acide]=n(acide)/V(acide)
n(acide)=[acide]×V(acide)
n(acide)=0.02×0.5
n(acide)=0.01 mol
