a. 0.00426 L
b. 1.596 L
The number of moles in dilution will remain constant
Can be formulated
[tex]\tt M_1.V_1=M_2.V_2[/tex]
a.
M₁=12
M₂=[tex]\tt 10^{-1.5}(pH=-log[H^+])[/tex]=0.032
V₂=1.6 L
[tex]\tt 12\times V_1=0.032\times 1.6\\\\V_1=0.00426~L[/tex]
b.
[tex]\tt 1.6-0.00426=1.596~L[/tex]