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0,1 x 50 ml
= 5mmol
MOL NH4CL = M X V
0,2x50 ml
=10mmol
NH3 + NH4CL -----> HCL + H2O
M: 10 10 10 10
R: 5 10 - -
S: 5 - 10 10
PH ditentukan oleh konsentrasi NH3 sisa
OH- = mol
v total
= 5 mmol
50mmol
= 0,1 M
pOH = - log ( OH-)
-log (0,1)
1 log 1
PH = 14-PH
14 (1-log1)
=13 + log 1