Question

Sodium hypochlorite, NaOCl, is the active ingredient in household bleach. What is the concentration of hypochlorite ion if 20.00 cm3 of bleach requires 25.30 cm3 of 0.500 M HCl to reach the equivalence point?

a. 1.13 M
b. 0.132 M
c. 0.395 M
d. 0.632 M

Answer 1

The concentration of the hypochlorite ion in the bleach solution, given that 20.00 cm³ of bleach requires 25.30 cm³ of 0.500 M HCl to reach the equivalence point, is 0.632 M.

Step-by-step explanation:

To determine the concentration of the hypochlorite ion, we need to use the titration information provided. The question states that 20.00 cm³ of bleach requires 25.30 cm³ of 0.500 M HCl to reach the equivalence point. The reaction between sodium hypochlorite and hydrochloric acid is as follows:

NaOCl (aq) + HCl (aq) → NaCl (aq) + H₂O (l) + Cl₂ (g)

For every mole of NaOCl that reacts, one mole of HCl is required. Since we know the volume and molarity of HCl, we can calculate the moles of HCl, and therefore the moles of NaOCl that reacted:

Moles of HCl = Molarity of HCl × Volume of HCl = 0.500 mol/L × 0.0253 L = 0.01265 mol

We can now calculate the concentration of NaOCl in the original bleach sample:

Ν NaOCl = Moles of HCl = 0.01265 mol

Volume of bleach = 20.00 cm³ = 0.02000 L

Concentration of NaOCl = Moles of NaOCl / Volume of bleach = 0.01265 mol / 0.02000 L = 0.632 M

Answer 2

Answer : The correct option is, (d) 0.632 M

Explanation : Given,

Concentration of HCl = 0.500 M

Volume of HCl =
`25.30cm^3=25.30mL=0.2530L`

conversion used :
`1cm^3=1mL\\1mL=0.001L`

Volume of hypochlorite ion =
`20.00cm^3=20.00mL=0.2000L`

The chemical reaction will be:

`OCl^-+H^+\rightarrow HClO`

First we have to calculate the moles of
`H^+` ion.

`\text{Moles of }H^+=\text{Concentration of }H^+* \text{Volume of }H^+`

`\text{Moles of }H^+=0.500M* 0.2530L=0.1265mol`

From the reaction we conclude that,

Moles of
`H^+` = Moles of
`ClO^-` = 0.1265 mol

Now we have to calculate the concentration of hypochlorite ion.

`\text{Moles of }ClO^-=\text{Concentration of }ClO^-* \text{Volume of }ClO^-`

`0.1265mol=\text{Concentration of }ClO^-* 0.2000L`

`\text{Concentration of }ClO^-=0.632M`

Therefore, the concentration of hypochlorite ion is 0.632 M.