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Sagot :
Answer:
The concentration of the unknown HCl is 0.0851 M.
Explanation:
The equation of neutralization:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
Where:
[tex]n_1[/tex]= Basicity of acid
[tex]n_2[/tex] = Acidity of base
[tex]M_1[/tex]= concentration of acid
[tex]M_2[/tex]= concentration of base
[tex]V_1[/tex] = Volume of acid used in neutralization
[tex]V_2[/tex] = Volume of base used in neutralization
We have:
The acidity of HCl = [tex]n_1=1[/tex]
The concentration of HCl solution used = [tex]M_1=?[/tex]
The volume of HCl used in titration =[tex]V_1= 15.00 mL[/tex]
The acidity of NaOH =[tex]n_2=1[/tex]
The concentration of NaOH solution used = [tex]M_2=0.0670 M[/tex]
The volume of NaOH used in titration =[tex]V_2= 19.06 mL[/tex]
[tex]n_1M_1V_1=n_2M_2V_2\\1\times M_1\times 15.00 mL=1\times 0.0670 M\times 19.06 mL\\M_1=\frac{1\times 0.0670 M\times 19.06 mL}{1\tines 15.00 mL}\\M_1=0.085135 M\approx 0.0851 M[/tex]
The concentration of the unknown HCl is 0.0851 M.
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