Activity 7 Acid-Base Equilibrium and the Acidity Constant PDF

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Activity 7: Acid-Base Equilibrium and the Acidity Constant 1. Hydrazine, N2H4, is a very weak base. Give the chemical equation demonstrating its reaction with water. N2H4 + H2O >> N2H5+ + OH2. Which of the following contains the highest H3O+ ion concentration? Explain your answer. 1. HCl 0.1 M 2. HCH3CO2 0.1 M HCl 0.1 M because hydrochloric acid is a strong acid, it completely dissociates in solution yielding a 0.1 M concentration of hydrogen ions (or hydronium ions) 3. A weak acid solution, HX 0.37 M, has a pH of 3.70. Calculate the acidity constant for this acid, M= 0.37 M pH= 3.7 -log(H) = pH -log(H)= 3.7 log (H)= -3.7 H=10

−3,7

Ka= H 2 / M - H = (10

−3.7

)

2

/ (0.37) - (10

−3.7

)

= 1.077 x 10

−7

The acidity constant is Ka = 1.077 x 10

−7

4. 50 g of benzoic acid is dissolved in 500 mL of water. Calculate the pH of the solution. Molar mass = 122.1 g/mol

50g/122.1= 0.41 mol 0.41 mol/0.5 L= 0.82M C6H5COOH ↔ C6H5COO Ka= (H (H

+

)=

+

−

+H

+

) 2 / (C6H5COOH)

√Ka

* ( C6H5COOH )

= √6.3 * 10 −5 * 0.82 =0.00718 M -log(0.00718) pH = 2.14 The pH is 2.14. 5. 0.25 mole of acetic acid is dissolved in enough water to make 1 L. What is the pH of the solution? CH3COOH ---> CH3C00 - + H+ At equilibrium = 0.25 mol of H+ [H+] = 0.25 mol/ 1L

= 0.25 mol/L

pH = -log [H+] = -log (0.25) = 0.6 The pH of the solution is 0.6 6. A weak acid solution of H2CO3 0.24 M has a pH of 3.49. Determine the value of Ka. pH = 3.49 [H+] = 10^-3.49 = 3.24x10^-4M

H2CO3 has 2 Ka values : Ka1 = [H+][HCO3-] / [H2CO3] and Ka2 = [H+][CO3 2-] / [HCO3-]

Ka = [3.24x10^-4][3.24x10^-4] / 0.24 = 4.37x10^-7 7. The poison injected by bee stings and red ants is made of formic acid (formic, derived from fourmis). Calculate the value of Ka if a formic acid solution of 0.10 M has a pH of 2.38.

HA ===> H+ + ApH = 2.38 [H+] = 1x10^-2.38 = 4.17x10^-3 M = [H+] = [A-]

Ka = [H+][A-]/[HA] Ka = (4.17x10^-3)(4.17x10^-3)/0.10 Ka = 1.74x10^-4

The Ka value is 1.74x10^-4 1. 200 g of CH3COOH are dissolved in enough water to make 1.5 L. What is the molarity of the acetic acid solution? CH3COOH = 60.05 g/mol 200g = 200/60.05 = 3.33 mol

3.33/1.5 = 2.22 M

Ka CH3COOH = 1.8*10^-5 Ka = [H+]²/[Acid] 1.8*10^-5 = [H+]² / 2.22 [H+]² = (1.8*10^-5)*2.22 [H+]² = 4*10^-5 [H+] = 6.32*10^-3 mol/L The molarity of the acetic acid solution is 6.32*10^-3M 2. What is the molarity of the hydronium ion? The hydronium ion molarity concentration is 1.0 x 10-7  mol/L 8. A solution of benzoic acid 0.25 M has 4.0 x 10-3 M of H3O+. Calculate Ka. C6H5COOH --> C6H5COO + H+ H30+ = H+ Ka= (4x10-3)^2 / 0.25 Ka= 6.4*10^-5 mol/L An ammonia solution has a concentration of 0.1 M. a.What is the Kb equation for this system? Kb = 1.75 × 10–5

b.Calculate the pH. -log(0.0013638) = 2.878 14 –2.878 = 11.122 .For the following acids: CH3COOH

HNO2 HCN What is the dissociation equation in an aqueous solution? a. CH3COOH+H2O ---> CH3COO-+H3O+ b. HNO2+H2O ---> NO2+H3O c. HCN+H2O ---> CN-+H3O+

a. K  a = [H3O+][ Ch3COO-]/[Ch3COOH] b. K  a = [H3O+][ NO2]\[HNO2] c. K  a = [H3O+][CN+] /[HCN] Identify the acid-base pairs. 

a.  CH3COOH and CH3COO, H2O  and H3O

b. H  NO2 and NO2,H2O and H3O c. H  CN and CN, H2O and H3O...