Question

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1. A solution was prepared by mixing 6.00 mL of 1.00M NaOH with 40.00 mL of...

1. A solution was prepared by mixing 6.00 mL of 1.00M NaOH with 40.00 mL of 0.800M weak acid and diluting to a total volume of 100.0 mL. The pH value of this solution was measured to be 3.12. Calculate the equilibrium hydrogen ion concentration of this solution using the pH value.

2. For the solution from question 1, calculate the number of moles of weak acid and calculate the number of moles of hydroxide ion that were originally present.

3. Calculate the number of moles of weak acid and calculate the number of moles of conjugate base present at equilibrium.

4. Calculate the concentrations of weak acid and conjugate base present at equilibrium.

5. Calculate the value of Ka for the unknown weak acid.

Solutions

Expert Solution

Q1.

Use the pH value given, 3.12

then

pH = -log([H+])

solve for [H+]

[H+] = 10^-pH = 10^-3.12 = 0.0007585 M

Q2.

find number of wak acid moles...

mmol of acid = Macid*Vacid = 0.8*40 = 32 mmol of acid, 32*10^-3 mol of acid present

then, find moles of OH- present originally

mmol of solution = Mbase*Vbase = 1*6 = 6 mmol of NaOH = 6*10^-3 molf of NaOH = 6*10^-3 mol of OH-

Q3.

find number of moles of weak acid, in equilibrium

mol of acid left after reaction = (32*10^-3 ) - (6*10^-3 ) = 0.026 mol of acid left

mol of conjugate base formed = mol of base added = 6*10^-3 = 0.006 mol of conjugate base in equilibrium

Q4.

concentration of each species

Vtotal = 100 mL = 0.1 L

[Weak acid] = 0.026/0.1 = 0.26 M

[Conjugate base] = 0.006/0.1 = 0.06 M

Q5.

use buffer equation to calculate the pKa

pH = pKa + log(conjugate base / weak acid)

3.12 = pKa+ log(0.06/0.26)

pKa = 3.12 - log(0.06/0.26)

pKa= 3.7568

Ka = 10^-pKa = 10^-3.7568 = 0.0001750 = 1.75*10^-4

queen_honey_blossom answered 1 year ago

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