A 0.050L solution of Ba (OH) 2 is neutralized by 0.072L of a 0.55 M HNO3 solution. What is the concentration of the Ba (OH) 2 solution?

0.396M

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2HNO3 + Ba (OH) 2 - > Ba (NO3) 2 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 2

The mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question. This include the following:

Volume of base (Vb) = 0.05L

Concentration of base (Cb) = ... ?

Volume of acid (Va) = 0.072L

Concentration of acid = 0.55 M

Step 3:

Determination of the concentration of the base, Ba (OH) 2.

The concentration of the base, Ba (OH) 2 can be obtained as follow:

Cava / CbVb = nA/nB

0.55 x 0.072 / Cb x 0.05 = 2/1

Cross multiply to express in linear form

Cb x 0.05 x 2 = 0.55 x 0.072

Divide both side by 0.05 x 2

Cb = (0.55 x 0.072) / (0.05 x 2)

Cb = 0.396M

Therefore, the concentration of the base, Ba (OH) 2 is 0.396M