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

`0.3456\ \text{M}`

Step-by-step 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=(M_1V_1)/(V_2)\\\Rightarrow M_2=(0.25* 34.56)/(25)\\\Rightarrow M_2=0.3456\ \text{M}`

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