If 24.83cm3 of 0.15moldm3 of NaoH is titrated to its end point with 39.45cm3 of HCl, what is the molarity of HCl
To find the molarity of HCl, we need to use the equation:
M_HCl x V_HCl = M_NaOH x V_NaOH
Where:
M_HCl is the molarity of HCl
V_HCl is the volume of HCl
M_NaOH is the molarity of NaOH
V_NaOH is the volume of NaOH
We can rearrange the equation to solve for M_HCl:
M_HCl = (M_NaOH x V_NaOH) / V_HCl
Given:
M_NaOH = 0.15 moldm^3 (molarity of NaOH)
V_NaOH = 24.83 cm^3 (volume of NaOH)
V_HCl = 39.45 cm^3 (volume of HCl)
M_HCl = (0.15 moldm^3 x 24.83 cm^3) / 39.45 cm^3
M_HCl = 0.09498 moldm^3
Thus, the molarity of HCl is approximately 0.095 moldm^3.