Question

Titration lab problems, assist with question 6 , then question1 on post lab

Answer 1

Answer:Molarity of (MOM ) = 0.048 M

Explanation :

In this reaction , HCL acts as an acid and MOM act as a base

Given :

• Acid Concentration HCL = 0.2 M

,

• Average volume of HCl = 24 mL

,

• Volume Base (MOM) = 100 mL

,

• Concentration (MOM) = ?M

We will calculate the concentration of MOM by applying the dilutions formula :

`M_{acid\text{ }}*\text{ V}_{acid\text{ }}=\text{ M}_{base\text{ }}*\text{ V}_{base\text{ }}`

Substituting the given parameters into the formula above, we calculate that Molarity (concentration) of MOM is :

`\begin{gathered} M_{base\text{ }}=\frac{M_(acid)*\text{ V}_{acid\text{ }}}{V_{base\text{ }}} \\ \text{ = }\frac{0.2\text{ M * 24 mL }}{100\text{ mL }} \\ \text{ =0.048 M } \end{gathered}`

This means that Molarity of (MOM ) = 0.048 M

Post lab question 1 :

Balanced reaction as follows :

`2HCl\text{ + Mg\lparen OH\rparen}_2\Rightarrow MgCl_2+2H2O\text{ }`