$25.0 mL$ of an HCl solution is titrated conductometrically with $0.120 M$ NaOH. The conductance readings (in mS) fit $G = 10.0 - 0.60 V$ before the end point and $G = 0.50 V - 3.2$ after it, where $V$ is the volume of NaOH added in mL. The concentration of the HCl solution, in $mmol L^{- 1}$ (one decimal), is

Numerical answer type, answer in mmol L^-1, tolerance 0.3.

Answer: 57.6 mmol L^-1

The end point is the intersection of the two straight lines:

10.0 - 0.60 V = 0.50 V - 3.2

, so

1.10 V = 13.2

and

V = 12.0 mL

. Moles of NaOH at the end point

= 0.120 \times 12.0 = 1.44 mmol

, equal to the moles of HCl in

25.0 mL

. Concentration

= 1.44/25.0 = 0.0576 M = 57.6 mmol L^{- 1}