Question

A volume of 500.0 mL500.0 mL of 0.140 M0.140 M NaOHNaOH is added to 585 mL585 mL of 0.200 M0.200 M weak acid (Ka=4.71×10−5).(Ka=4.71×10−5). What is the pHpH of the resulting buffer?

Answer 1

pH = 4.50

Step-by-step explanation:

A weak acid HX reacts with NaOH as follows:

HX + NaOH → H2O + NaX

Initial moles of HX are:

0.585L * (0.200mol / L) = 0.117 moles of HX

The moles of NaOH are moles of NaX that are produced, that means:

0.500L * (0.140mol / L) = 0.07 moles of NaOH = 0.070 moles of NaX produced.

And moles of HX that still in solution are:

0.117 moles - 0.070 moles = 0.047 moles of HX

To find the pH of this buffer (Mixture of a weak acid, HX with its conjugate base, NaX) we use H-H equation:

pH = pKa + log [NaX] / [HX]

Where pKa is -log Ka = 4.327

And [] could be taken as moles of each species.

Replacing:

pH = 4.327 + log [0.070] / [0.047]

pH = 4.50