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

2HCl + Ba(OH)2 → BaCl2 + 2H2O

2 Answers

1 vote

Answer:

0.0202L

Step-by-step 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

User Nullrevolution
by
4.0k points
3 votes

Answer:


V_(HCl)=5.05x10^(-3)L

Step-by-step 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)=(M_(Ba(OH)_2)V_(Ba(OH)_2))/(2M_(HCl)) =(0.101M*0.050L)/(2*0.50M) \\\\V_(HCl)=5.05x10^(-3)L

Besr regards.

User Sihad Begovic
by
4.4k points