What volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

9.85mL

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

Ca(OH)2 + 2HNO3 —> Ca(NO3)2 + 2H2O

From the balanced equation above,

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) =?

Mb (Molarity of base) = 0.155 M

Va (volume of the acid) = 28.8 mL

Ma (Molarity of acid) = 0.106 M

Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i.e the base) can be obtain as follow:

MaVa/MbVb = nA/nB

0.106 x 28.8 / 0.155 x Vb = 2/1

Cross multiply to express in linear form as shown below:

2 x 0.155 x Vb = 0.106 x 28.8

Divide both side by 2 x 0.155

Vb = (0.106 x 28.8) / (2 x 0.155)

Vb = 9.85mL

Therefore, the Volume of calcium hydroxide is 9.85mL

b. accuracy of the data matters alot more than its number of digits, organization or amount of detail. quantitative observation is more accepted than the qualitative observation.