Răspuns:
-calculez concentratia molara
-M,Ba(OH)2=171g/mol
niu=m/M= 1.71/171 mol= 0,01mol
c,m= niu/Vs UNDE Vs= ms/ro-=(m,apa+md)/ro=181,71g/1,075g/ml
V,s=169 ml= 0,169 l
c,m=0,01mol/0,169 l=0,06 mol/l= 6x10⁻²mol/l
-calculez pH
pH+pOH=14---> pH= 14-pOH
pOH= -log[HO⁻]
dar, considerand ca Ba(OH)2 ionizeaza total, [HO⁻]=[Ba(OH)2]
pOH= -log6x10⁻²=2-log6
pH=14-2+log6=12+log6
-calculez c%
c=mdx100/ms= 1,71x100/181,71=>0,94%
Explicație:
