50.00 mL of 0.100 M HCl is titrated with 0.100 M NaOH at 298 K ( $K_{w} = 1.0 \times 1 0^{- 14}$ ). Calculate the change in pH produced by raising the volume of NaOH added from 49.95 mL to 50.05 mL (that is, the pH jump across the end point). Give the answer to two decimal places.

Numerical answer type, tolerance 0.02.

Answer: 5.4

At 49.95 mL, unreacted HCl

= 0.05 \times 0.100 = 5.0 \times 1 0^{- 3}

mmol in 99.95 mL:

[H X^{+}] = 5.003 e - 05

M, pH

= 4.30

. At 50.05 mL, excess NaOH gives

[OH X^{-}] = 4.998 e - 05

M, pOH

= 4.30

, pH

= 9.70

. The jump is

Δ pH = 9.70 - 4.30 = 5.40

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