In relation to titration curves, in what pH range (acidic, neutral, or basic) wo

Question

In relation to titration curves, in what pH range (acidic, neutral, or basic) would one expect the equivalence point to exists. b. Add 10.0 mL of 0.10 M HCI (025.0 mL of 0.10 M NH3. C. Add 25.0 mL of 0.05 M NaOH to 50.0 mL of 0.05 M acetic acid. 2. A 0.268 g sample of monoprotic acid neutralizes 16.4 ml of 0.08 M KOH solution. Calculate the molar mass of the acid. (Molar mass = g/mol) What is the percent ionization of a 0.250 M solution of formic acid (HCOH. monoprotic acid) with Ka = 1.8 x 10^-4? What is the percent ionization of 0.01 M solution of pyridine (C5H5N) with pH 8.59 (Kb = 1.7 x 10^-9)?

Explanation / Answer

Hi,

1 b) The equivalence point will be in the acidic range because the titration is between strong acid (HCl) Vs weak base (NH3)

1 c) The equivalence point will be in the basic range as the titration is between a strong base (NaOH) and weak acid (CH3COOH)

2) Moles of KOH = 0.08M X 0.0164L = 0.001312 moles

Since this completely neutralizes the monoprotic acid

Therefore, it can be said that moles of KOH = moles of monoprotic acid

Moles of acid = mass of acid / molar mass of acid

Therefore, molar mass of acid = 0.268 g/ 0.001312moles = 204.27 g/mol

3) Formic acid weakly ionises in solution as

CH3COOH---> CH3COO- + H+

Therefore, ionisation constant, Ka is given by

Ka = [CH3COO-] [H+] / [CH3COOH]

Let [CH3COO-] = [H+] = x

1.8 *10^-4 = x*x/ 0.25-x

0.45*10^-4 - 1.8 *10^-4x = x^2

x^2 + 1.8 *10^-4x - 0.45 *10^-4x = 0

x = 0.021 or x = -0.021

x cannot be negative, therefore x =[H+] = 0.021M

Percent Ionization of 0.25 M Formic acid = (0.021/ 0.25) * 100 = 8.4 %

4) This problem can be solved in two ways

one can calculate [OH-] as we have calculated for the above problem

or

Since pH is given as 8.59

pOH = 14 - 8.59 = 5.41

[OH-] = 3.89 X 10^-6

Therefore, percent ionization of 0.01M pyridine = (3.89 X 10^-6 / 0.01) X 100 = 0.0389%

Hope your queries are answered

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