HCl –este un acid tare
deaceea
[H3O+]=CM
[H3O+]=0,1=10
la minus 1
pH- puterea luata cu semn schimbat=1sau daca esti a 12
a =-log[H3O+]
Cm=n/Vs
10 mL=10la minus 2LITRI
n=CmxVs
n=1x10 la minus2
Solutia finala are concentratia egala
Cm=n/Vs
Cm=10 la minus2/5x10la
minus 2
Cm=1/5=0,2
Cm=[H3O+]
pH=-log[H3O+]=-(-0,70)=0,7
