Question

A 45.0 ml sample of nitric acid is neutralized by 119.4 ml of 0.2M NaOH. What is the molarity of the nitric acid?

Answer 1

0.53L

Explanations:

According to dilution formula:

`C_1V_1=C_2V_2`

C1 and C2 are the initial and final concentrations

V1 and V2 is initial and final volume

Given the following parameters

• V1 = 45mL

,

• C2 = 0.2M

,

• V2 = 119.4mL

Substitute the given parameters into the formula

`\begin{gathered} C_1=(C_2V_2)/(V_1) \\ C_1=(0.2*119.4)/(45) \\ C_1=(23.88)/(45) \\ C_1=0.53M \end{gathered}`

Hence the molarity of the nitric acid is 0.53L