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What is the pH in a titration of 125 mL of 0.45 F sulfurous acid (H2SO3)...

What is the pH in a titration of 125 mL of 0.45 F sulfurous acid (H2SO3) with a 1.23 M solution of sodium hydroxide before any base is added?

Solutions

Expert Solution

Volume of H2SO3 = 125 mL = 0.125 L

Molarity of H2SO3 = 0.45 M

Moles of H2SO3 = Molarity x volume

=0.45 M x 0.125 L

=0.05625 mol

Ka of H2SO3 = 1.2 x 10-2

H2SO3 H2O <===> H3O+ HSO3-
Initial 0.05625 mol - 0 0
Change -x - +x +x
equilibrium 0.05625-x - x x

By solving the above quadratic equation

using formula

We get x = 0.0207

Concentration of H3O+ = 0.0207 M

pH = -log[H3O+]

pH = -log[0.0207]

pH = -(-1.68)

pH = 1.68

queen_honey_blossom answered 10 months ago
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