Answer
a) 6 mL HCl
b) 8 mL HCl
Work Step by Step
$C_M$ = $\frac{n}{V}$
In order to neutralize the solutions, the moles of each base need to be consumed by HCl
a)
$n_{NaOH}$ = 0.3$\times$0.01 = 0.003 moles
HCl + NaOH = NaCl + $H_2O$
1 mole HCl............1 mole NaOH
x moles HCl..........0.003 moles NaOH
=> x = $\frac{0.003\times1}{1}$ = 0.003 moles HCl
V = $\frac{n}{C_M}$ = $\frac{0.003}{0.5}$ = 0.006 L = 6 mL HCl
b)
$n_{Ba(OH)_2}$ = 0.2$\times$0.01 = 0.002 moles
2HCl + $Ba(OH)_2$ = $BaCl_2$ + 2$H_2O$
2 moles HCl.............1 mole $Ba(OH)_2$
y moles HCl.............0.002 mole $Ba(OH)_2$
y = $\frac{0.002*2}{1}$ = 0.004 moles HCl
V = $\frac{n}{C_M}$ = $\frac{0.004}{0.5}$ = 0.008 L = 8 mL HCl
