A titration of 200.0 mL of 1.00 M H2A was done with 1.38 M NaOH. For the diprotic acid H2A, Ka1 = 2.5  10–5, Ka2 = 3.1  10–9. Calculate the pH after 100.0 mL of 1.38 M NaOH have been added.

4.95

Explanation:

1.00 M H2A

1.38 m NaOH

Titration = 200.0 mL

Calculate moles of NaOH

=  = 0.46

calculate moles of H2A

= = 0.667

therefore the moles of acid left = moles of H2A - moles of NaOH

                                                    = 0.667 - 0.46 = 0.207

pka = - log( ka )

       = - log ( 2.5 * 10^-5 ) = 4.61

calculate PH after 100 ml of 1.38 M NaOH have been added

PH = pka + log

     = 4.61 + log = 4.95

answer: the rate of diffusion increases with increase in temperature.

increase in temperature, increases the molecular/particles speed (kinetic energy. diffusion means movement of particles through the material.

generally, the heavier the particle the more it interacts with the environment and more the temperature matters. so higher temperature results in higher diffusion.

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