Gọi:
CM NaOH= x (M)
CM Ba(OH)2 = y (M)
nHCl = 0.09*0.1=0.009 mol
OH- + H+ --> H2O
0.009_0.009
Ta có :
0.075x + 0.075y*2 = 0.009
<=> x + 2y = 0.12 (1)
nBaCO3 = 0.2955/197=0.0015 mol
Ba2+ + CO32- ---> BaCO3
=> nBa2+ = 0.0015 mol
<=> 0.075y = 0.0015
<=> y = 0.02
Thay y vào (1) :
x = 0.08
Vậy:
CM NaOH = 0.08 M
CM Ba(OH)2 = 0.02 M