Question

How many Liters of 0.50M HCl are needed to neutralize 0.050L of 0.101M Ba(OH)2?

Answer 1

Answer:

$V_{HCl}=5.05x10^{-3}L$

Explanation:

Hello,

In this case, since hydrochloric acid and barium hydroxide are in a 2:1 molar ratio, for the neutralization, the following moles equality must be obeyed:

$2*n_{HCl}=n_{Ba(OH)_2}$

In such a way, in terms of molarities and volumes, we can compute the required volume of hydrochloric acid as shown below:

$2*M_{HCl}V_{HCl}=M_{Ba(OH)_2}V_{Ba(OH)_2}\\\\V_{HCl}=\frac{M_{Ba(OH)_2}V_{Ba(OH)_2}}{2M_{HCl}} =\frac{0.101M*0.050L}{2*0.50M} \\\\V_{HCl}=5.05x10^{-3}L$

Besr regards.

Answer 2

Answer:

0.0202L

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

2HCl + Ba(OH)2 → BaCl2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Next, the data obtained from the question. This includes the following:

Molarity of acid (Ma) = 0.5M

Volume of acid (Va) =...?

Volume of base (Vb) = 0.050L

Molarity of base (Mb) = 0.101M

The volume of the acid, HCl needed for the reaction can be obtained as follow:

MaVa/MbVb = nA/nB

0.5 x Va / 0.101 x 0.05 = 2/1

Cross multiply to express in linear form

0.5 x Va = 0.101 x 0.05 x 2

Divide both side by 0.5

Va = (0.101 x 0.05 x 2)/0.5

Va = 0.0202L

Therefore, the volume of the acid, HCl needed for the reaction is 0.0202L