This question has multiple parts. Work all the parts to get the most points. Cal

Question

This question has multiple parts. Work all the parts to get the most points. Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a 100.0 mL of0.70 MHC_7H_s O_2 (K_a = 6.4 times 10%-5 titrated by 0.10 M NaOH pH at the halfway point = pH at the equivalence point = b 100.0 ml. of 0.70 M C_2H_5NH_2(K_b = 5.6 times 10^-4) titrated by 0.40 M HNO_3 pH at the halfway point = pH at the equivalence point = c 100.0 mL of 0.80 M HCl titrated by 0.15 M NaOH pH at the halfway point = pH at the equivalence point =

Explanation / Answer

a)Ka of acid = 6.4x10-5  and pKa = 4.1938

At half equivalence point the acid is half neutrlaised to give its conjugate base.

Thus [acid] = [conjugate base] and the solution behaves as an acidic buffer.

The pH of buffer is given by Hendersen equation

pH = pka + log [conjugate base]/[acid]

at half equivalence

pH = pKa = 4.1938

At equivalence

mmoles of acid = mmoles of base

100mL x 0.7M = VmL x 0.1M

thus volume of base = 700mL

and concentration of salt formed = [salt] = 100x0.7 /800= 0.0875 M

the pH of a salt of weak acid and strong base is given by

pH = 1/2[ pKw + pKa + log C]

= 1/2 [14 + 4.1938 + log 0.0875]

= 8.567

b) kb of base = 5.6x10-4 and pKb = 3.25

By similar analogy at the half equivalence point

pOH = pKb and pH = 14 - pOh

pOH = 3.25 and pH = 10.75

At equivalence mmoles of base = mmoles of acid

100mLx 0.7M = V x 0.4

and volume of acid = 175 mL

and concentraion of salt formed = [salt] = 70/275 = 0.2545 M

and pH of a salt of weka base and strong acid is given by

pH = 1/2 [pKw -pKb - log C]

= 1/2 [ 14 -3.25 - log 0.2545]

= 5.672

c) HCl + NaOH

At half equivalence

1/2 mmoles of acid = mmoles of base used

1/2 x 100mLx0.80 M = V x 0.15M

or volume of NaOh used at half equivalence = 266.67 mL

The solution has now 40mmol of acid in 366.67mL of solution

Thus [H+] = 40/366.67 = 0.1499 M

pH at equivalence = - log [H+] = -log 0.1499

= 0.8239

As the salt formed is of a strong acid and strong base, the salt is neutral .

pH at equivalence = 7.00

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