CHEM Problem: Dissociation constant of acids and bases: Please help me calculate

Question

CHEM Problem: Dissociation constant of acids and bases: Please help me calculate the followings (bold). Please include calculation by steps. Thank you.

Determination of the Dissiciation Constant of Phosphoric Acid.

Questions:

Based on the today’s experiment describe a process to identify an unknown, weak acid. How would you determine the number of acidic groups present? What calculations should be done?

List possible sources of errors in both Parts A and B. State how each error would effect the calculated Ka (part A) and pH (part B) values.

CH3COOH + NaOH Molarity of NaOH (M) 0.107 Volume of NaOH (mL) 4.0 pH (First Reading) 3.57 pH (Second Reading) 3.55 Average pH 3.56 CH3COOH initially present (moles) CH3COOinitially present (moles) NaOH present @ Equilibrium (moles) Concentration NaOH @ Equilibrium (M) CH3COOH present @ Equilibrium (moles) Concentration CH3COOH @ Equilibrium (M) CH3COO- present @ Equilibrium (moles) Concentration of CH3COO- @ Equilibrium (M) Concentration of H+ @ Equilibrium (M) Ka

Explanation / Answer

The no. of mmol of NaOH = 0.107 mmol/mL * 4 mL = 0.428 mmol

The pKa value of CH3COOH = 4.74

According to Henderson-Hasselbulch equation: pH = pKa + Log(nCH3COONa/nCH3COOH), where nCH3COONa = nNaOH and 'n' corresponds to no. of mmol

ai.e. 3.56 = 4.74 + Log(0.428/nCH3COOH)

i.e. Log(nCH3COOH/0.428) = 4.74 - 3.56 = 1.18

i.e. nCH3COOH/0.428 = 101.18 = 15.1356

i.e. nCH3COOH = 0.428*15.1356 = 6.478 mmol

Now, pKa = -Log(Ka) = 4.74

i.e. Ka = 10-4.74 = 1.82*10-5

The table can be filled as follows.

CH3COOH + NaOH Molarity of NaOH (M) 0.107 Volume of NaOH (mL) 4.0 pH (First Reading) 3.57 pH (Second Reading) 3.55 Average pH 3.56 CH3COOH initially present (moles) (6.478+0.428)*10-3 = 6.906*10-3 CH3COOinitially present (moles) 0 NaOH present @ Equilibrium (moles) 0 Concentration NaOH @ Equilibrium (M) 0 CH3COOH present @ Equilibrium (moles) 6.478*10-3 Concentration CH3COOH @ Equilibrium (M) 6.478/4 = 1.6195 CH3COO- present @ Equilibrium (moles) 4.28*10-4 Concentration of CH3COO- @ Equilibrium (M) 0.428/4 = 0.107 Concentration of H+ @ Equilibrium (M) 10-3.56 = 2.75*10-4 Ka 1.82*10-5

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