Question

2. Calculate (Ca²+], [OH ], and [H30*] for a solution that is prepared by dissolving 0.600 grams of Ca(OH)2(s) in enough water to make 1.00 liters of solution at 25°C. 3. Calculate the pH of an aqueous solution that is 0.020 M in HNO3(aq) at 25°C. Is the solution acidic or basic? 4. Calculate the pH and the pOH of an aqueous solution prepared by dissolving 2.0 grams of KOH(s) pellets in water and diluting to a final volume of 0.500 liters at 25°C.

Answer 1

2)

mass of Ca(OH)2 = 0.600 grams

volume of the solution = 1.00L

Molarity of the solution = number of moles of solute / volume in L.

molar mass of Ca(OH)2 = 74.09 g/mole

number of moles of Ca(OH)2 = mass of solute/molar mass

number of moles of Ca(OH)2 = 0.600 / 74.09 = 0.0081 moles

Molarity of Ca(OH)2 = 0.0081 / 1.00 = 0.0081 M

Concentration of Ca(OH)2 = 0.0081M

Ca(OH)2 ---------------------- Ca+2(aq) + 2 OH-(aq)

Concentration of Ca+2 = [Ca+2] = 0.0081M

Concentration of OH- = [OH-] = 2 x 0.0081 = 0.0162M

[OH-] = 0.0162M

[H3O+] [OH-] = Kw        where Kw = ioinc product of water = 1.0 x10^-14

[H3O+] = Kw / [OH-] = 1.0 x10^-14 / 0.0162 = 61.73 x10^-14

[H3O+ ] = 6.173 x10^-13M

3)

HNO3 = 0.020M

concentration of HNO3 = 0.020M

HNO3 is a strong acid

HNO3 ------------------- H+ + OH-

[H+ ] = 0.020M

-log[H+] = -log( 0.020)

PH = 1.698

PH = 1.70

PH is lower than 7

Hence, it is acidic solution

4)

mass of KOH = 2.0 grams

volume of the solution = 0.500L

molar mass of KOH = 56.0 g/mole

number of moles of KOH = mass of solute/molar mass = 2.0 / 56.0 = 0.0357 moles

number of moles of KOH = 0.0357 moles

Concentration of KOH = number of moles / volume in L = 0.0357 / 0.5 = 0.0714 M

Concentration of KOH = 0.0714M

KOH ------------------ K+ + OH-

                                         0.0714

[OH-] = 0.0714M

-log[OH-] = -log(0.0714)

POH = 1.146

POH = 1.15

but

PH + POH = 14

PH = 14 - POH

PH = 14 - 1.15

PH = 12.85