$50.0 mL$ of $0.0100 M$ HCl is titrated with $0.100 M$ NaOH. Take limiting molar ionic conductivities (in $S c m^{2} mo l^{- 1}$ ): $H X^{+} = 350$ , $Na X^{+} = 50$ , $Cl X^{-} = 76$ , $OH X^{-} = 198$ . Ignore the change in volume on adding NaOH and assume the ionic conductivities do not change with concentration. The conductivity of the solution after adding $2.50 mL$ of NaOH, in $mS c m^{- 1}$ (two decimals), is

Numerical answer type, answer in mS cm^-1, tolerance 0.03.

Answer: 2.76 mS cm^-1

Initial HCl

= 50.0 \times 0.0100 = 0.500 mmol

; NaOH added

= 2.50 \times 0.100 = 0.250 mmol

, so half the acid is neutralised. In 50.0 mL:

[H X^{+}] = 0.250/50.0 = 0.00500 M

[Na X^{+}] = 0.00500 M

[Cl X^{-}] = 0.0100 M

[OH X^{-}] \approx 0

κ = \frac{\sum λ _{i} c _{i}}{1000} = \frac{350 ( 0.00500 ) + 50 ( 0.00500 ) + 76 ( 0.0100 )}{1000} = \frac{1.75 + 0.25 + 0.76}{1000} = 2.76 \times 1 0^{- 3} S c m^{- 1} = 2.76 mS c m^{- 1}

. (The initial value was

4.26 mS c m^{- 1}