Question

(1) A weak acid, HA, has a K_a of 6.5 x 10^-6 and a concentration of 0.25 M. 25.5 mL of this weak acid is titrated with 0.10 M NaOH. What is the pH of the solution at the equivalence point?

Report your answer with 3 decimal places.

(2) The concentration of 16.6 mL of HCl is determined by titrating with NaOH. The titration reaches its endpoint after adding 18.4 mL of 0.829 M NaOH. What is the concentration of the HCl?

Report your answer in M with three decimal places.

Answer 1

(1)

A weak acid is being titrated with a strong base(OH^-) to equivalence point.

Hence, the reaction that happens can be written as

H A(aq) + OH(aq) + Aaq) + H20 (1)

Note that at equivalence point the number of moles of strong acid added is equal to the number of moles of weak acid present.

Concentration of the weak acid = 0.25 M = 0.25 mol/L

Volume of weak acid being titrated = 25.5 mL = 0.0255 L

Hence, the number of moles of acid being titrated is

TH4 = 0.25 - “ x 0.0255 L = 0.006375 mol

Hence, the number of moles of OH^- added must also be 0.006375 mol.

Now, given that the concentration of NaOH = 0.10 M = 0.10 mol/L, the volume of NaOH added can be calculated as

moles volume = conc 0.006375 mol · = 0.06375 L = 63.75 mL 0.10 mol/L

Now, we can create an ICF table for the reaction that happens.

		$OH^-$	$A^-$	$H_2O$
Initial, mol	0.006375	0.006375	0	0
Change, mol	-0.006375	-0.006375	+0.006375	+0.006375
Final, mol	0	0	0.006375	0.006375

Hence, the final situation is that we have 0.006375 mol of the conjugate base of our weak acid, which will undergo hydrolysis with water as follows:

Aag) + H200) = H A(ag) + OH

Now, we can calculate the K_b for the conjugate base of HA as follows:

Кья К. К K, 1.0 x 10-14 6.5 х 10-6 1.538 х 10-9

Now, the initial concentration of A^- can be calculated by dividing the number of moles by total volume.

Total volume = 25.5 mL (acid) + 63.75 mol (base) = 89.25 mL = 0.08925 L

Hence, the concentration can be calculated as

[A-]= 0.006375 mol - 0,08925 L 0.0714 M

Now, we can create an ICE table to find the equilibrium OH^- concentration.

	$[A^-]$		ОН-
Initial, M	0.0714	0	0
Change, M	-x	+x	+x
Equilibrium, M	0.0714-x	x	x

Hence, we can write the expression of Kb for the above hydrolysis using the ICE table as follows:

K -= 1.538 x 10-4 [HAllон = ххх [A-] 0.0714 - 1.048 x 10-5 M.

Hence, the equilibrium OH^- concentration of the solution is .

r = 1.048 x 10-5 M

Now, pOH of the solution can be calculated as follows:

poh = – log[OH-] = – log (1.048 x 10-5) 4.9795

Hence, the pH of the solution can be calculated as

pH + po H = 14 pH = 14 - pOH = 14 - 4.9795 9.020

Hence, the pH of the solution is approximately 9.020.

(2)

HCl (a strong acid) is being titrated with NaOH (a strong base).

At the end point of titration the number of moles of NaOH and HCl will be equal as predicted by the neutralization reaction below.

HCl(aq) + NaOH(aq) + NaCl(aq) + H200

Now, the volume of NaOH added = 18.4 mL = 0.0184 L

Concentration of NaOH = 0.829 M = 0.829 mol/L

Hence, the number of moles of NaOH added is

nNaOH = concx volume = 0.829 **** - x 0.0184 L=0.0152536 mol

Hence, the number of moles of HCl also must be 0.0152536 mol.

Volume of HCl added = 16.6 mL = 0.0166 L

Hence, the concentration of HCl can be calculated as follows:

conc= mol Volume 0.0152536 mol - 0.0166 L 0.919 M

Hence, the concentration of HCl is approximately 0.919 M.