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A 25.0 mL sample of 0. 150 M acetic acid is titrated with a 0.150 M NaOH solutio

ID: 969470 • Letter: A

Question

A 25.0 mL sample of 0. 150 M acetic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point The K_a of acetic acid is 4.50 * 10^-4 11.74 4.74 9.26 7.00 8.81 What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 ml of aqueous hydrofluoric acid requires 30.00 ml. of 0.300 M NaOH K_a = 6.76 * 10^-4 for 1 HF. 8.19 12.19 5.81 1.81 What is the molar solubility of silver chloride (AgCl) in water The solubility-product constant for AgCl is 1.8 * 10^-10 at 25 degree C. 9.0 * 10^-11 1.9 * 10^-5 1.3 * 10^-5 3.6 * 10^-10 9.74

Explanation / Answer

Q.20: Moles of CH3COOH in the sample = MxV = 0.150 mol/L x 0.025 L = 0.00375 mol.

At equivalence point 0.00375 mol of CH3COOH will reacct with 0.00375 mol NaOH to form 0.00375 mol of CH3COONa.

Hence volume of NaOH need to be added = 0.00375 mol / 0.150 mol/L = 25 mL

Total volume of the solution at equivalence point, Vt = 25 mL + 25 mL = 0.050 L

Hence [CH3COONa] = 0.00375 mol / 0.050 L = 0.075 M

At equivalence point all of the CH3COOH is converted to CH3COONa.

At equivalence point CH3COONa undergoes hydrolysis to form CH3COOH and the pH can be calculated from salt hydrolysis formulae

pH = (1/2)x[pKw + pKa + log[CH3COONa] )

=> pH = (1/2)x[14 + 4.75 + log0.075] = 8.81

Hence option (E) is correct.

Here Ka is actually equals to 1.78x10-5

Q.21: moles of the salt NaF formed = MxV = 0.300 mol/L x 0.030 L = 0.009 mol

Total volume of solution, Vt = 25+30 = 55 mL = 0.055 L

Hence [NaF] = 0.009 mol / 0.055 L = 0.164 M

At equivalence point NaF undergoes hydrolysis to form HF and the pH can be calculated from salt hydrolysis formulae

pH = (1/2)x[pKw + pKa + log[CH3COONa] )

=> pH = (1/2)x[14 + 3.17 + log0.164] = 8.19 (answer)

Hence option (A) is correct.

Q.22: AgCl(s) + H2O ------ > Ag+(aq) + Cl-(aq)

---------------------------------------- S, ----------- S

Ksp = 1.8x10-10 = [Ag+(aq)]x[Cl-(aq)] = S2

=> S = underroot(1.8x10-10) = 1.3x10-5M (answer)

Hence option (C) is correct

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