Question

Titration Volume & Concentration

a) What volume of a 0.304 M hydrochloric acid solution is required to neutralize 17.1 mL of a 0.161 M barium hydroxide solution?

_____ mL hydrochloric acid

Answer 1

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

Step-by-step 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*(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)*(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*(1L)/(0,304mol)` = 0,0181L≡18,1mL

I hope it helps!