Question

2.) 100.0 mL of a 0.100 M solution of HCN (K,=4.9 10-19) is titrated with a 0.200 M solution of KOH. Calculate the pH of solution iii) Before any addition of KOH solution. (2pts) After the addition of 10.0 mL of KOH solution. (2pts) At the half-equivalence point. (1pt) At the equivalence point. (2pts) After the addition of 100 mL of KOH solution. (1pt)

Answer 1

2.

i. Before any addition of KOH solution, the pH of the solution of HCN (a weak acid) is calculated from the formula:

15601 – "ydly = Hd

where pKa = - logKa = - log (4.9 x 10-10) = 9.31

and C = initial concentration of HCN = 0.100 M

Therefore, $pH= \frac {1}{2}[9.3-log\,(0.100\,M)] = \frac {1}{2}[9.3+1] =5.15$ (answer)

The formulation of the formula (for a generic weak acid HA) is shown below:

o duc For a weak acid, HA the following equilibrium produces the Htion. . HA HATA Ii C 0 0 ..: xc ac ac E: c-dc xe xe Where, &= degree of dissociation of the weak & c= initial concentration

Now, from Ostwald's dilution principle .: pHE-log Style = -log (oc) = -log() . c) => pH = -log ſkaic = -/ log ka looge pH = 1 {ka - loge)

ii. The reaction occuring during the titration is a acid-base reaction which is as follows:

$HCN(aq)\,+\,KOH(aq)\;\rightarrow \;KCN(aq)\,+\,H_2O(l)$

Now, during the titration, KOH (a strong base) reacts weak acid, HCN to form its salt KCN. Therefore, the resulting solution is a buffer solution: a solution of a weak acid (HCN) and its salt with a strong base (KOH).

The pH of a buffer solution is calculated by the Henderson-Hasselbach equation which is as follows:

$pH = pK_a + log\, \frac {[Salt]}{[Weak\,acid]} = pK_a + log\, \frac {[KCN]}{[HCN]}= 9.31 + log\, \frac {[KCN]}{[HCN]}$

Now, 100 mL 0.100 M HCN solution contains $\frac {100}{1000} L \times 0.100\,mol/L = 0.01\,mol\,\,HCN$

10 mL = (10/ 1000) L of 0.200 M KOH has been added to the HCN solution.

Therefore, 10 mL of 0.200 M KOH has $\frac {10}{1000} L \times 0.200\,mol/L = 0.002\,mol\,\,KOH$

This 0.002 mol KOH reacts with 0.002 mol of HCN to form 0.002 mol KCN. [Since according to the reaction 1 mole KOH reacts with 1 mole HCN]

Therefore, remaining HCN = (0.01 mol - 0.002 mol) = 0.008 mol

Total volume = 100 mL acid + 10 mL added KOH = 110 mL

Therefore, $[KCN] =\frac {0.002\,mol}{\frac {110}{1000} L} = 0.0182\,mol/L$

$[HCN]=\frac {0.008\,mol}{\frac {110}{1000} L} = 0.0727\,mol/L$

Therefore, $pH= 9.31 + log\, \frac {0.0182\,mol/L}{0.0727\,mol/L} = 9.31 - 0.60 =8.71$ (answer)

iii. At half-equivalence point the solution is still a buffer solution (mixture of weak acid and its salt). Now, the special feature of this half-equivalence point is that exactly half of the acid has been converted to salt and other half remains as acid. Therefore, [KCN] = [HCN]

Now, the Henderson-Hasselbach equation reduces to

$pH= 9.31 + log\, \frac {[KCN]}{[HCN]} = 9.31 + log\, \frac {[HCN]}{[HCN]} = 9.31 + log\,1$

or,

iv. At equivalence point all of the acid, HCN has been converted to KCN salt. Now, the salt KCN undergo hydrolysis in the aqueous solution:

$KCN(aq)\,+\,H_2O(l) \;\rightarrow \; HCN(aq)\,+\,KOH(aq)$

Now, the pH for this hydrolysis of salt is calculated by the formula:

$pH = 7 + \frac {1}{2}pK_a +\frac {1}{2}log\,C$

or, $pH = 7 + (\frac {1}{2}\times9.31) +\frac {1}{2}log\,C$

or, $pH = 11.655 + \frac {1}{2}log\,C$

Now, all of the moles of HCN in the 100 mL (= 0.1 L) 0.100 M HCN solution has been completely converted to salt, KCN.

Therefore, moles of KCN present in solution = 0.01 mol

Now, to titration 0.01 mol of HCN we have to add 0.01 mol of KOH from outside.

Now, 1000 mL (i.e. 1 L) 0.200 M KOH solution contains 0.200 mol KOH.

Therefore, 0.01 mol KOH is present in $= \frac {1\,L}{(0.200\,mol)} \times 0.01\,mol = 0.05\,L$

Therefore, the total volume of the solution = 0.1 L + 0.05 L = 0.15 L

Therefore, concentration of salt, KCN = C = (moles of salt)/ (total volume of solution)

= (0.01 mol)/ (0.15 L)

= 0.067 mol/L

Now, $pH = 11.655 + \frac {1}{2}log\,C = 11.655 + \frac {1}{2}log\,(0.067\,mol/L) = 11.655 - 1.176$

or, (answer)

v. 100 mL = 0.1 L of 0.200 M KOH solution contains (0.200 mol/L)* (0.1 L) = 0.02 mol of KOH.

Now, the solution of HCN contains only 0.01 mol of HCN which has already reacted with 0.01 mol of KOH.

Therefore, reamaining KOH in the solution = 0.02 mol - 0.01 mol = 0.01 mol KOH.

Total volume = 100 mL HCN + 100 mL KOH added = 200 mL = 0.2 L

Now, since KOH is a strong base, it will be completely dissociated to give 0.01 mol of each K+ and OH- .

Therefore, the concentration of OH- = [OH- ] = (moles of OH- )/ (total volume of solution)

= (0.01 mol)/ (0.2 L)

= 0.05 mol/L

Therefore, pOH = - log [OH-] = -log (0.05 mol/L) = 1.30

Therefore, pH = 14 - pOH = 14 - 1.30 = 12.70 (answer)