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Chapter 17: Problem 91

A sample of 0.1687 gof an unknown monoprotic acid was dissolved in 25.0 mL. of water and titrated with 0.1150 \(\mathrm{M}\) NaOH. The acid required 15.5 \(\mathrm{mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After 7.25 mL of base had been added in the titration, the pH was found to be \(2.85 .\) What is the \(K_{a}\) for the unknown acid?

Short Answer

Expert verified
The molar mass of the unknown monoprotic acid is 94.61 g/mol and its acidity constant (\(K_a\)) is 0.001239.

Step by step solution

01

Determine the moles of NaOH at the equivalence point

At the equivalence point, the number of moles of the unknown acid (HA) is equal to the moles of the sodium hydroxide (NaOH) used to titrate. We can find the moles of NaOH used for the titration using the formula: moles = Molarity × Volume (in Liters). Given that the volume of NaOH required to reach the equivalence point is 15.5 mL, we can first convert it to Liters as follows: Volume of NaOH = 15.5 mL × (1 L / 1000 mL) = 0.0155 L Now let's find the moles of NaOH: Moles of NaOH = 0.1150 M × 0.0155 L Moles of NaOH = 0.0017825 mol
02

Determine the molar mass of the unknown acid

As the moles of the unknown acid (HA) are equal to the moles of NaOH, we have 0.0017825 mol of HA. To determine the molar mass of the unknown acid, divide the mass of the acid sample by the moles of acid: Molar mass of HA = (0.1687 g) / (0.0017825 mol) = 94.61 g/mol The molar mass of the unknown monoprotic acid is 94.61 g/mol.
03

Determine the moles of HA and NaOH at the given pH

We are given that after 7.25 mL of 0.1150 M NaOH has been added in the titration, the pH is 2.85. First, let us determine the moles of HA and NaOH at this point. Volume of NaOH = 7.25 mL × (1 L / 1000 mL) = 0.00725 L Moles of NaOH = 0.1150 M × 0.00725 L = 0.00083375 mol Since we started with 0.0017825 mol of HA, the moles of HA remaining after the addition of 0.00083375 mol of NaOH would be: Moles of HA remaining = 0.0017825 - 0.00083375 = 0.00094875 mol.
04

Calculate the [HA] and [A-] concentrations in the solution

We need to find the concentration of the acid (HA) and its conjugate base (A-) at the given pH point. To do this, we will divide the moles of HA and its conjugate base by the total volume of the solution at this point. The total volume of the solution after adding 7.25 mL of NaOH to 25 mL of water is 0.03225 L (32.25 mL). [HA] = moles of HA remaining / total volume of the solution = 0.00094875 mol / 0.03225 L = 0.029429 M The moles of A- in the solution are equal to the moles of NaOH, as NaOH turns HA into its conjugate base A-: [A-] = moles of NaOH / total volume of the solution = 0.00083375 mol / 0.03225 L = 0.025852 M
05

Use the given pH to find the \(K_a\) for the unknown acid

Given the pH = 2.85, we can convert this to the hydrogen ion concentration using the formula \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\). [\(\mathrm{H}^{+}\)] = \(10^{-2.85}\) = 0.001413 M At this point in the titration, the primary source of the hydrogen ion concentration is the reaction between HA and A-. We can write the expression for the acidity constant \(K_a\): \(K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]}\) Now plug in the values for [\(\mathrm{H}^{+}\)], [A-], and [HA] to find the \(K_a\): \(K_a = \frac{(0.001413)(0.025852)}{0.029429}\) = \(0.001239\) The acidity constant (\(K_a\)) for the unknown monoprotic acid is 0.001239.

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