Question

3) 500.0 mL of 0.110 M NaOH is added to 605 mL of 0.200 M weak acid (Ka = 2.35

Answer 1

Here a buffer solution is created due to mixing of strong base NaOH with weak acid HA. The following acid -base neutralization reaction occurs leading to the formation of a buffer solution of an weak acid and its salt of a strong base.

(Weak acid) HA(aq)  + OH^-(aq) ------------------>   H₂O(l) + A^- (salt of weak acid)

Also after the addition of NaOH and weak acid (HA), their concentration will decrease due to increase in volume.

After mixing total volume of the solution, Vt = 500mL+605mL = 1105mL = 1.105L

concentration of NaOH after mixing = (moles of NaOH)/Vt = 0.5x0.110/1.105 = 0.050M

Final concentration of weak acid(HA) after mixing = (moles of HA)/Vt = 0.605x0.200/1.105 = 0.11M

Now the concentration after neutralisation reaction can be calculated as

HA(aq)   + OH^-(aq) ------------------>   H₂O(l) + A^-   

at t= 0, 0.11M 0.050M

after neutralisation (0.11-0.050) 0 0.050M

Hence [HA] = (0.11-0.050) = 0.06M , [A-] = 0.050M

Now pH buffer solution can be calcuated from Henderson's equation

pH = pKa + log[salt]/[acid]

= - log(2.35x10^-5) + log(0.050)/(0.060) = 4.55 (answer)