1B) Calculate the percent ionization of 0.0040 M weak acid HA in a solution cont

Question

1B) Calculate the percent ionization of 0.0040 M weak acid HA in a solution containing 0.070 M of the salt NaA

2) 50 mL of unknown concentration of HNO2 is titrated with 33.9 mL of 0.0523 M KOH. What is the unknown concentration of the original HNO2 solution?

3) How many milliliters of 0.165 M HCl are needed to titrate each of the following solutions to the equivalence point?

(3a) 41.0 mL of 0.0950 M NaOH
mL
(3b) 40.7 mL of 0.128 M KOH
mL
(3c) 50.0 mL of a solution that contains 1.75 g of NaOH per liter

4) Calculate the hydronium ion concentration and the pH at the equivalence point when 65.0 mL of 0.4000 M NH3 is mixed with 35.0 mL of 0.7429 M HCl.
Ka=5.6x10-10
[H3O+]=?

ph=?

5) Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 5.00 102 mL of solution and then titrate the solution with 0.138 M NaOH.

What are the concentrations of the following ions at the equivalence point?

Na+, H3O+, OH-
C6H5CO2-

? M Na+
? M H3O+
? M OH-
? M C6H5CO2-
What is the pH of the solution?

6) Construct a rough plot of pH versus volume of added base for the titration of 50 mL of 0.060 M HCN with 0.075 M NaOH. Ka = 4.0*10-10 for HCN

(a) What is the pH before any NaOH is added?

(b) What volume of base, in milliliters, is required to reach the equivalence point?

(c) What is the pH at the equivalence point?

Explanation / Answer

Too many questions in the same post. I reccomend you to post few questions in a post so you can get answered. I'll answer 3 of them. The others post them in another question thread and per separate.

1. writting the reaction:
r: HA + H2O --------> H3O+ + A-
i: 0.004 0 0
e: 0.004-x x x

1.5x10-5 = x2/0.004-x --> Ka is small, so we can neglect 0.004-x to 0.004 only:
1.5x10-5 * 0.004 = x2
x = [H3O+] = 2.45x10-4 M

%I = 2.45x10-4 / 0.004 * 100 = 6.12%

2. Use the expression: MAVA = MBVB and solve for MA:
MA = 0.0523 * 33.9/50 = 0.0355 m

3. To the different cases, we have the following:
a. Va = 0.095 * 41 / 0.165 = 23.60 mL
b. Va = 0.128 * 40.7 / 0.165 = 31.57 mL
c. Calculate the concentration first:
1.75 g/L * 1mol/(23+1+16)g = 0.0438 M

Va = 0.0438 * 50 / 0.165 = 13.26 mL

Hope this helps

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