A 28.8 mL sample of HCl is titrated with NaOH. If 39.5 mL of 0.639 M NaOH is needed to reach the endpoint, what is the concentration (M) of the HCl solution? NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

Concentration of HCl=0.88M

Explanation:

It is a acid base reaction that is neutralisation reaction since it complete reaction so,

Mili equivalent of HCl=Mili equivalent of NaOH

and,

Mili Equivalent = Normality * volume

N1*V1=N2*V2

for acids,

Normality=molarity*basicity

Basicity=number of H+ ion given by acid into the solution and

for HCL

basiciy=1;

N1=1*M1=M1

for base,

Normality=molarity*Acidity

Basicity=number of OH- ion given by base into the solution and

for NaOH

acidity=1;

N2=M2*1=0.639*1=0.639;

N1*28.8=0.639*39.5

N1=0.876

M1=N1/Basicity;

M1=0.876/1;

M1=0.876

M1=0.88

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