Question

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.

Answer 1

4.95

Step-by-step explanation:

1.00 M H2A

1.38 m NaOH

Titration = 200.0 mL

Calculate moles of NaOH

=
`(100*1.38)/(300)` = 0.46

calculate moles of H2A

=
`(200 * 1.0)/(300)` = 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
`((salt)/(acid) )`

= 4.61 + log
`((0.46)/(0.207) )` = 4.95