Question

5) (4) If the pH of 1.000 L of H2SO4 is 2.1377, what is the [H*]? What would the final pH be if 0.22644 g of KOH were added? (Assume no change in volume.) Show the reaction between these 2 compounds. 3) (4) The pH of a solution of 0.1500 M 2,6-Dinitrophenol (acid form, like [HA]) is 1.293. If you determine that the phenolate (anion form of this weak acid, like [A-]) is 2.30 x 10-3 M, what is the pK, of 2,6-DNP? What would the new pH be if added 3.17 x 10-4 mole of the phenolate anion in 1.000 L of this solution (assume no volume change)?

Answer 1

5. Sol :-

Given, pH of H₂SO₄ = 2.1377

Because,

pH = - log₁₀[H⁺]

So,

[H⁺] = 10^-pH

[H⁺] = 10^-2.1377

[H⁺] = 0.007283 M

Hence, [H+] = 0.007283 M

After the addition of KOH :-

Molarity of H₂SO₄ = [H⁺]/2 = 0.00728 M /2 = 0.00364 M

= 0.00364 mol M

Because, Molarity = Number of moles / Volume of solution in L

So,

Number of moles of H₂SO₄ = 0.00364 mol M x 1.000 L

= 0.00364 mol mol

Also, Number of moles = Given mass in g / Gram molar mass

So,

Number of moles of KOH = 0.22644 g / 56.1056 g/mol

= 0.00404 mol

Reaction between H₂SO₄ and KOH is :

H₂SO₄ (aq) + 2KOH (aq) -------------------> K₂SO₄ (aq) + 2H₂O (l)

ICF table is :

....................H₂SO₄ (aq).................. +............. 2KOH (aq) ---------> K₂SO₄ (aq)............ +............ 2H₂O (l)

Initial (I).........0.00364 mol mol ................................0.00404 mol...........0.0 mol............................................

Change (C)......-1/2x0.00404 mol......................-0.00404 mol...........+1/2x0.00404 mol.........................

Final (F)............0.00162 mol.............................0.0 mol........................0.00202 mol.............................

So,

Moles of H₂SO₄ left = 0.00162 mol

Molarity of H₂SO₄ = Moles/Volume in L

= 0.00162 mol / 1.00 L

= 0.00162 M

Also, [H⁺] = 2 x 0.00162 M = 0.00324 M

pH = - log[H⁺]

= - log 0.00324 M

= 2.4895

Hence, pH after the addition of KOH = 2.4895