Question

Titration Volume & Concentration

Answer 1

Answer:

18,1 mL of a 0,304M HCl solution.

Explanation:

The neutralization reaction of Ba(OH)₂ with HCl is:

2 HCl + Ba(OH)₂ → BaCl₂ + 2 H₂O

The moles of 17,1 mL≡0,0171L of a 0,161M Ba(OH)₂ solution are:

$0,0171L*\frac{0,161moles}{L}$ = 2,7531x10⁻³moles of Ba(OH)₂

By the neutralization reaction you can see that 2 moles of HCl reacts with 1 mole of Ba(OH)₂. For a complete reaction of 2,7531x10⁻³moles of Ba(OH)₂ you need:

$2,7531x10^{-3}molBa(OH)_{2}*\frac{2molHCl}{1molBa(OH)_{2}}$ = 5,5062x10⁻³moles of HCl.

The volume of a 0,304M HCl solution for a complete neutralization is:

$5,5062x10^{-3}molHCl*\frac{1L}{0,304mol}$ = 0,0181L≡18,1mL

I hope it helps!