Question

ACID/BASE PRACTICE PROBLEMS

(a) 3 moles of sodium nitrate are placed in 1 L of water. Calculate the pH.

(b) Calculate the Ka and the % ionization of a weak acid if 0.02 moles of the acid is added to 1 L water and makes a solution with a pH of 4.75.

(c) A solution containing 0.2 moles of NaOH, 0.3 moles of HCl, and 0.1 moles of HF (Ka=7.1e-4) is added to 1 L of water. Calculate the pH and pOH.

(d) 0.2 moles of the salt acid of urea is added to 1 L of water, calculate the pH of the solution.

Answer 1

(a) NaNO3 is a salt of a strong acid (HNO3) and a strong base (NaOH). We know that strong acid reacts with a strong base to give a neutral solution. Hence, the pH of this solution is 7.

	NaNO3	--->	Na+	+	NO3-
moles	3		3		3
Volume of solution (L)	1		1		1
Concentration(mole/L)	3		3		3

(b) Let the weak acid be HA and initial concentration of HA = 0.02 mole/1L = 0.02 mol/L = 0.02 M

	HA	<=>	H+	+	A-
Initial concentration	0.02		-		-
At equilibrium	0.02(1-)		0.02		0.02

Ka = [H+][A-]/[HA] ----(1)

Given,

pH = 4.75 => pH = -log[H+] => [H+] = 10-pH

[H+] = 10-4.75 = 1.78 x 10-5 M

And [H+] = [A-] = 0.02 = 1.78 x 10-5 M .........(2)

From the equation (1),

Ka = (1.78 x 10-5 M x 1.78 x 10-5 M)/(0.02 M) = 1.58x10-8  

Ka = 1.58x10-8

To find the % ionization, (x100 = degree of dissociationx100)

0.02 => = 1.78 x 10-5/0.02 = 8.89 x 10-4

% ionization of weak acid = x100 = 8.89 x 10-4 x 100 = 0.0889 ~ 0.09 %

Hence, % ionization of weak acid is 0.09.

(c) As we know NaOH and HCl are the strong acid and HF is a weak acid. 0.2 mol of NaOH reacts with 0.2 mol of HCl to give 0.2 mol of NaCl salt.

After the above neutralization reaction, 0.1 mol of HCl left (0.3 mol HCl - 0.2 mol NaOH). Now, the system has 0.1 mol HCl and 0.1 mol HF.

As we know the concentration of H+ from HF is much lower compared to HCl. Hence, we have to calculate the pH of the solution from H+ of HCl.

[H+] = 0.1 M

pH = -log[H+] = -log(0.1) = 1 => pH = 1

pOH = 14 - pH = 14 - 1 = 13 => pOH = 13

(d) Salt acid of urea dissolve in water to form OH- ion in the solution. Which means it form the solution basic.

So, the 0.2 moles of the salt acid of urea is added to 1 L of water has a concentration = 0.2 mol/L = 0.2 M

Here, [OH-] = 0.2 M

pOH = -log[OH-] = -log(0.2) = 0.699

pH = 14 - pOH = 13.3

Hence, pH of salt acid of urea is 13.3