What volume of a 0.181 M nitric acid solution is required to neutralize 25.1 mL of a 0.167 M calcium hydroxide solution

46.32 mL

Explanation:

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

2HNO₃ + Ca(OH)₂ —> Ca(NO₃)₂ + 2H₂O

From the balanced equation above,

The following data were obtained:

Mole ratio of the acid, HNO₃ (nA) = 2

Mole ratio of the base, Ca(OH)₂ (nB) = 1

Next, we shall write out the data obtained from the question. This includes:

Molarity of the acid, HNO₃ (Ma) = 0.181 M

Volume of the base, Ca(OH)₂ (Vb) = 25.1 mL

Molarity of the base, Ca(OH)₂ (Mb) = 0.167 M

Volume of the acid, HNO₃ (Va) =?

Finally, we shall determine the volume of the acid, HNO₃ as shown below:

MaVa/ MbVb = nA/nB

0.181 × Va / 0.167 × 25.1 = 2/1

0.181 × Va / 4.1917 = 2

Cross multiply

0.181 × Va = 4.1917 × 2

0.181 × Va = 8.3834

Divide both side by 0.181

Va = 8.3834 / 0.181

Va = 46.32 mL

Therefore, the nitric acid, HNO₃ required for the reaction is 46.32 mL

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.