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v1=20cm3=0,02dm3
Cm1=0,10mol/dm3
n=Cm1*v1=0,02dm3*0,1mol/dm3= 0,002mol
Cm2=0,05mol/dm3
v2=?
n2=?
pH=2 => [H+]=0,01 (mol/dm3)
H3O+ + OH- -> 2H2O (H+ + OH- -> H2O)
Równanie : [H+]=n1-n2/v1+v2
n2=Cm2*v2
[H+]=n1-Cm2*v2/v1+v2
0,01=0,002-0,05*v2/0,02+v2
0,0002+0,01v2=0,002-0,05v2
0,06v2=0,0018
v2=0,03dm3=30cm3 - objętośc roztworu NaOH jaką należy dodać.