Alcl3+3NaoH=Al(oH)3!+3Nacl
1mol 3mol 78
0.025 X y
X=0.075 y=1.95
Al(oH)3+NaoH=NaAlO2+2H2O
78 1
(1.95-1.56) Z
Z=0.005 c(NaOH)=(0.005+0.075)/0.05=1.6mol/L