Answer : The correct option is, (d) 0.632 M
Explanation : Given,
Concentration of HCl = 0.500 M
Volume of HCl = [tex]25.30cm^3=25.30mL=0.2530L[/tex]
conversion used : [tex]1cm^3=1mL\\1mL=0.001L[/tex]
Volume of hypochlorite ion = [tex]20.00cm^3=20.00mL=0.2000L[/tex]
The chemical reaction will be:
[tex]OCl^-+H^+\rightarrow HClO[/tex]
First we have to calculate the moles of [tex]H^+[/tex] ion.
[tex]\text{Moles of }H^+=\text{Concentration of }H^+\times \text{Volume of }H^+[/tex]
[tex]\text{Moles of }H^+=0.500M\times 0.2530L=0.1265mol[/tex]
From the reaction we conclude that,
Moles of [tex]H^+[/tex] = Moles of [tex]ClO^-[/tex] = 0.1265 mol
Now we have to calculate the concentration of hypochlorite ion.
[tex]\text{Moles of }ClO^-=\text{Concentration of }ClO^-\times \text{Volume of }ClO^-[/tex]
[tex]0.1265mol=\text{Concentration of }ClO^-\times 0.2000L[/tex]
[tex]\text{Concentration of }ClO^-=0.632M[/tex]
Therefore, the concentration of hypochlorite ion is 0.632 M.
The concentration of the base from the calculation is 0.395 M.
The reaction is written as;
NaOCl (l) + 2 HCl (aq) → NaCl (aq) + Cl 2 (g) + H 2O (l).
Using the formula;
CAVA/CBVB = NA/NB
CA = 0.500 M
VA = 25.30 cm3
CB = ?
VB = 20.00 cm3
NA = 2
NB = 1
CAVANB = CBVBNA
CB = CAVANB /VBNA
CB = 0.500 × 25.30 × 1/ 20.00 × 2
CB = 0.395 M
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