Question

QUESTION 1 · 1 POINT *This is a part of the 6 part titration practice problem. The titration set up is as follows: 35.0 mL of 0.125 M HCN is titrated with 0.200 M KOH. The pka of HCN is 9.21. What is the pH once half of the equivalence volume has been added? Select the correct answer below: 11.34 9.21 O 7.00 10.22 Content attribution QUESTION 7 - 1 POINT

These are all part of the same problem. please show steps for each question and explain.

Answer 1

The reaction between weak acid HCN and strong base KOH is

HCN + KOH ------------> KCN + H2O

This equation indicates that 1 mole of KOH will neutralize 1 mole of HCN

Here the HCN i a weak acid and KCN is the salt of its conjugate base.

Now any solution where both the weak acid and salt of its conjugate base is present is called a buffer. And pH of a buffer is calculated by Henderson Hasselbalch equation

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

Here

pH = pKa + log [KCN]/[HCN]

a-

Half of the equivalance point that that point where half of the weak acid is completely converted to its salt of conjugate base. Thus at this point, [KCN] = [HCN]

Now putting this in-

pH = pKa + log [KCN]/[HCN]

pH = pKa + log 1

pH = pKa + 0

pH = pKa

Thus at half of equivalence point, pH = pKa of HCN = 9.21

b-

Equivalence point is that point where the moles of weak acid present = moles of strong base added. i.e at this point all the acids present have been completely neutralized by the base added.

Given HCN taken = 0.125 M and 35 ml

Thus moles of HCN taken = concentration * volume

= 0.125 M * 35 ml

= 0.125 mol/1000 ml * 35 ml

= 0.004375‬ mols

Since from equation 1 we know 1 mole of KOH will neutralize 1 mole of HCN

Then total moles of KOH required to neutralize 0.004375‬ mols of HCN =  0.004375‬ mols

That means volume of KOH required = mols of KOH / concentration of KOH

= 0.004375‬ mols / 0.200 M

= 0.004375‬ mols / (0.200 mol/ 1000 ml)

= 21.875 ml

c-

Given volume of KOH added (V1) = 10 ml

Initial concentration of KOH (M1) = 0.200 M

Initial volume of HCN taken (V1) = 35 ml

Initial concentration of HCN taken (M1) = 0.125 M

After addition of the two, fnal volume (V2) = 10 ml + 35 ml = 45 ml

Thus new Initial concentration of KOH (M2) = M1V1 / V2 = 0.200 M * 10 ml / 45 ml = 0.044 M

Similarly new Initial concentration of HCN (M2) = M1V1 / V2 = 0.125 M * 35 ml / 45 ml = 0.0972 M

Thus the final concentration of acid and its salt after reaction is

Reaction	HCN	KOH	KCN
Initial	0.0972 M	0.044 M	0
Change	- 0.044 M	-0.044 M	+0.044 M
Equilibrium	0.0532‬	0	0.044 M

Hence pH = pKa + log [KCN]/[HCN]

  pH = 9.21 + log [0.044]/[0.0532]

= 9.21 + log [0.827]

= 9.21 + [-0.072]

= 9.138‬

= 9.14

d- pH at equivalence point is

at equivalance point,

volume of KOH added (V1) = 21.875 ml

Initial concentration of KOH (M1) = 0.200 M

Initial volume of HCN taken (V1) = 35 ml

Initial concentration of HCN taken (M1) = 0.125 M

After addition of the two, fnal volume (V2) = 21.875 ml + 35 ml = 56.875 ml

Thus new Initial concentration of KOH (M2) = M1V1 / V2 = 0.200 M * 21.875 ml / 56.875 ml = 0.0769 M

Similarly new Initial concentration of HCN (M2) = M1V1 / V2 = 0.125 M * 35 ml / 56.875 ml = 0.0769 M

Thus the final concentration of acid and its salt after reaction is

Reaction	HCN	KOH	KCN
Initial	0.0769 M	0.0769 M	0
Change	- 0.0769 M	-0.0769 M	+0.0769 M
Equilibrium	0	0	0.0769 M

Now here [KCN] = 0.0769 M

Here the salt will again dissociate to form the conjugate acid as-

KCN + H2O ---------> HCN + OH-

Here Kb for KCN = [HCN] * [OH-] / [KCN]

We know Kb for KCN = 2.0 X 10^-5

Thus forming ICE table-

Reaction	KCN	HCN	OH-
Initial	0.0769 M	0	0
Change	- x	+x	+x
Equilibrium	0.0769 M - x	+x	+x

Thus Kb = [HCN] * [OH-] / [KCN]

2.0 X 10^-5 = [x] * [x] / [0.0769 M - x]

2.0 X 10^-5 [0.0769 M - x] = x²  

(0.1538 * 10^-5 ) - (2.0 X 10^-5) x = x²  

x² + (2.0 X 10^-5) x - (0.1538 * 10^-5 ) = 0

solving this

x = 0.0012

Thus here [OH-] = x = 0.0012 M

Thus pOH = -log [OH-] = -log [0.0012] = 2.9

Hence pH = 14- pOH = 14- 2.9 = 11.1