Question

50.0 mL sample of the weak acid

the concentration of the weak acid = 0.15 M

25 mL of the week acid into 100 mL beaker

titrated this solution of 0.21 M NaOH

moles of weak acid = 3.75*10^-3

moles of NaOH = moles of week acid

c) How many milliliters of the NaOH are required to neutralize the sample of weak acid? d) How many moles of NaOH have been added at one half of the volume in part 'C' (volume at the half equivalence point)? c) How many moles of the weak acid have reacted at the hall equivalence point? f.) Calculate the pH of the solution at the hall equivalence point. g) Explain how the pH at the half equivalence point is related to Ka for the weak acid. CHEM 1515-YEAR 2019-2020 9-4

Answer 1

(c)

moles of weak acid = 3.75*10^-3

moles of NaOH = moles of weak acid

moles of NaOH required = 3.75*10^-3 moles

NaOH is 0.21 M

ml. of NaOH required : = 3.75*10^-3 moles *1000 ml/L  / (0.21 mol /L ) =  17.86 ml

(d) When volume half of equivalence point is added :

Half of volume in part 'C' =8.93 ml

so, moles of NaOH added =( 8.93 ml / 1000 ml/L)* 0.21 mol /L = 1.88*10^-3 moles

(e)

we have, HA (aq) + NaOH (aq) $\rightleftharpoons$ NaA (aq) + H₂O (l)

Moles HA = moles NaOH

At half of equivalence point :

Half of volume of weak acid reacted = 12.5 ml

so, moles weak acid reacted   =(12.5 ml / 1000 ml/L)* 0.15 mol /L = 1.88*10^-3 moles

(f)

pH of solution at equivalence point :

At the equivalence point, moles of weak acid initially present and the moles of NaOH added are identical. Since their reaction effectively proceeds to completion, predominate ion in solution is A^-, which is a weak base.

To calculate the pH we first determine the concentration of A^-,

( A^-) = moles of weak acid added / total volume =  0.15 M * 25.0 ml / 42.86 ml = 0.0875 M

A^- is a weak base, Kb for it = Kw / Ka = 10^-14 / (1.8*10^-5) = 5.56*10^-10

A^-(aq) + H₂O (l)  $\rightleftharpoons$  HA (aq) + OH^- (aq)

Kb = (HA ) (OH^- ) /    ( A^-)
So, [OH^–] = (Kb*   ( A^-) ) ^1/2 =(5.56*10^-10*  0.0875 ) ^1/2 = 6.97*10^-6 M

pH = 14 -pOH = 14 - (-log[6.97*10^-6 ] = 8.85

(g)

pH at half equivalence point is related to Ka value. of weak acid.

before equivalence point,

HA (aq) + OH^- (aq) $\rightleftharpoons$ A^-(aq) + H₂O (l)

Any such solution containing comparable amounts of a weak acid HA, and its conjugate
weak base, A^–, is act as buffer. So, pH can be calculated following equation :

the Henderson–Hasselbalch equation :

pH = pKa + log ( ( A^-) / (HA ) )