Question

1. Sulfanilic acid (NH2C6H4SO3H) has a pKa of 3.23. How many grams of Na(NH2C6H4SO3) [182.1 g/mol] would you have to dissolve in 125 mL of water to get a pH of 7.97?

a. 0.59

b. 1.2g

c. 3.5g

d. 4.8g

2. When KNO2 is dissolved in water, ___________.

a. [H3O+] = [OH-]
b. [H3O+] < [OH-]
c. [H3O+] > [OH-]

3. Calculate the concentration of HCN in a sol’n prepared by dissolving 10.8 g of NaCN (49 g/mol) in enough water to make 500.0 mL of sol’n.  Ka for HCN = 4.0 x 10-10

1. 		0.0057 M
   		0.0033 M
   		0.41 M
   		0.15 M

4. What is the pH of a sodium formate (68 g/mol) solution prepared by adding 0.680 g of sodium formate to 100.0 mL of water at 25°C? The Ka for formic acid is 1.8 x 10-4 at 25°C.

	2.38
		5.63
		8.37
		11.62

5. Of the compounds below, a 0.10 M aqueous solution of ________ will have the highest pH.

		KCN, Ka of HCN = 4.0 x 10-10
		NH4NO3, Kb of NH3 = 1.8 x 10-5
		NaClO, Ka of HClO = 3.2 x 10-8
		NaHS, Ka of H2S = 5.6 x 10-8
		NaC2H3O2, Ka of HC2H3O2 = 1.8 x 10-5

Answer 1