19. NEED HELP FAST

Question

A volume of  500.0 mL  of  0.100 M   NaOH  is added to  535 mL  of  0.200 M  weak acid  (𝐾a=2.96×10−5).  What is the  pH  of the resulting buffer?

HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)

pH=

Answer 1

Let HA be the weak acid.

HA + NaOH ==> NaA + H2O

moles HA = 0.535 L x 0.200 mol/L = 0.107 moles

moles NaOH added = 0.5 L x 0.1 mol/L = 0.05 moles

Set up an ICE table....

HA + NaOH ==> NaA + H2O

0.107....0.05.................0............Initial

-0.05....-0.05..............+0.05.......Change

0.057.....0...................0.05........Equilibrium

pH = pKa + log [NaA]/[HA] and pKa = -log Ka = -log 2.96x10^-5 = 4.53

pH = 4.53 + log (0.05/0.057)

pH = 4.53 + (-0.058)

pH = 4.47