A 250.0 mL sample of 0.200 M aqueous aluminum chloride is mixed with 100.0 mL of 0.200 M aqueous sodium hydroxide.
I need help calculating the concentration of Na...
Check these concentrations:
OH= 0M since is the LR
Al= 0.124M
Na=??
Cl= 0.429M
(Na^+) = (NaOH)
mols NaOH = M x L
(NaOH) = (M NaOH x L NaOH/total volume) = 0.200 x 0.100/0.350) =?
The others look ok to me.