$50.0 mL$ of $0.0100 M$ HCl is titrated with $0.100 M$ NaOH. Limiting molar ionic conductivities (in $S c m^{2} mo l^{- 1}$ ): $H X^{+} = 350$ , $Na X^{+} = 50$ , $Cl X^{-} = 76$ , $OH X^{-} = 198$ . Ignoring the volume change and the concentration dependence of ionic conductivities, the conductivity of the solution after adding $7.50 mL$ of NaOH is closest to

Answer: A

HCl present

= 0.500 mmol

; NaOH added

= 7.50 \times 0.100 = 0.750 mmol

, so the end point (5.00 mL) is passed and

0.250 mmol

OH X^{-}

is in excess. In 50.0 mL:

[Na X^{+}] = 0.750/50.0 = 0.0150 M

[Cl X^{-}] = 0.0100 M

[OH X^{-}] = 0.00500 M

[H X^{+}] \approx 0

κ = \frac{50 ( 0.0150 ) + 76 ( 0.0100 ) + 198 ( 0.00500 )}{1000} = \frac{0.75 + 0.76 + 0.99}{1000} = 2.50 \times 1 0^{- 3} S c m^{- 1} = 2.50 mS c m^{- 1}

. Forgetting the excess

OH X^{-}

gives

1.51

; the value at the end point itself (NaCl only) is

1.26