bonjour besoin d’aide pour les questions c) et d) svp

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