A50.0 ml solution of 0.194 m sulfurous acid ( h2so3 ) is titrated with 0.194 m naoh . the pka values of sulfurous acid are 1.857 ( pka1 ) and 7.172 ( pka2 ). calculate the ph at the first equivalence point. ph = calculate the ph at the second equivalence point. ph =

ph at first equivalency point= 4.51, ph at second equivalency point= 10

Explanation:

ph at first equivalency point is calculated by finding the mid- point of pka1 and pka2, hence;

ph=(pka1 + pka2)/2 = (1.857 +7.172)/2 = 4.51

At second equivalency point SO₃²⁻ is ionized, as shown:

SO₃²⁻ + H₂O ⇄ HSO₃⁻ + OH⁻

Moles of H₂SO₄ neutralized = Volume x Molarity = 0.05 x 0.194 = 0.0097 moles = Moles of SO₃⁻ produced

Second equivalency point means that 1 mole of acid reacts with 2 moles of base, hence

Moles of NaOH required to neutralize SO₃⁻ = 2 x 0.0097 = 0.0194 moles

Volume of NaOH required = No. of moles / Molarity = 0.0194/0.194= 0.1 L

Total Volume of Solution at Second equivalency point = 0.05 + 0.1= 0.15L

Concentration of SO₃⁻ at second equivalency point = No. of moles/ Volume = 0.0097/0.15= 0.065 M

SO₃⁻ is acting as a weak base, thus

Kb= [HSO₃⁻][OH⁻]/[SO₃⁻]

pKb= 14 - pka2 = 14 - 7.127 = 6.828

Kb = 10^ -6.828 = 1.49 x 10⁻⁷

[HSO₃⁻] = x = [OH⁻]

1.49 x 10^⁻⁷ = x²/0.065

x = 9.84 x 10⁻⁵ = [OH⁻]

p[OH⁻] = -㏒[OH⁻] = 4.007≅4

ph = 14 - p[OH⁻] = 14 - 4 = 10

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

i think that the answer is c