Question

A solution of HCl with a volume of 25.00 mL is titrated to the endpoint, with 0.250 M
NaOH. If it takes 34.56 mL of NaOH, what is the original concentration of HCl in the
solution?
HCl(aq) + NaOH(aq) → H20(l)+ NaCl(aq)

Answer 1

Answer:

$0.3456\ \text{M}$

Explanation:

$V_1$ = Volume of NaOH = 34.56 mL

$V_2$ = Volume of HCl = 25 mL

$M_1$ = Concentration of NaOH = 0.25 M

$M_2$ = Concentration of HCl

When endpoint is reached the number of moles of NaOH will be equal to the number of moles of HCl

$M_1V_1=M_2V_2\\\Rightarrow M_2=\dfrac{M_1V_1}{V_2}\\\Rightarrow M_2=\dfrac{0.25\times 34.56}{25}\\\Rightarrow M_2=0.3456\ \text{M}$

Concentration of HCl is $0.3456\ \text{M}$.