Question

Lab 8 Acid Base Properties of Solutions Lab Report Worksheet

In today’s lab we are going to utilize a simulation to determine the pH of various salt solutions. We will then use those pH values to determine concentrations of hydronium and Hydroxide in solutions. From there we will determine the K_a or K_b

1. Pre-Work: In the table below, use just the salt formula to identify the salt as acidic, basic, or neutral.

Table 1: Acidic, Basic, or Neutral Salt Predictions

Salt	Acidic, Basic or Neutral?
NaCl
NaF
NaCN
NaClO
NaNO2
NH4Cl
NH4F
NH4CN
C5H5NH+CN-

2. Open the simulation by clicking on the following link http://employees.oneonta.edu/viningwj/sims/hydrolysis_s.html

This simulation requires that flash be enabled. You will have to tell your web browser to allow flash. For google chrome that is done by going into your browser setting, go to site settings and make sure the option for flash is set to “ask first”. You then have to click on the simulation screen and tell your browser to allow flash. You may see a pop up asking if you want it to allow flash or you might have to click on the flash icon.

3. In the simulation
   1. Click on the appropriate ion combinations (1 cation and 1 anion) to form the salts listed in the table below.
   2. The default concentration for the salt solution is 0.010 M. Leave this setting as is for the time being
   3. Read the pH measurement for each ion set/salt and record in the attached data table
   4. Convert pH to [H₃O⁺] and record in the attached data table
   5. Convert [H₃O⁺]to [OH^-] and record in the attached data table
   6. For salts that were acidic, determine the Ka, for salts that are basic, determine Kb. Show your work for one of the salts. You do not have to do this for all salts. You will need to show the dissociation of the salt, identification of hydrolysis, ICE table, [H₃O⁺], [OH^-], and Ka or Kb in the area below the table.

For salts that have both ions undergoing hydrolysis, both a Ka and Kb can be calculated. For this exercise we will only calculate one based on the stronger ion. If a basic pH was determined, base you K calculation on the hydrolysis of the basic anion equation. If an acidic pH was determined, base your K calculation on they hydrolysis of the acidic cation equation.

0.010 M Salt Solution	pH	[H3O+]	[OH-]	Ka or Kb
NaCl
NaF
NaCN
NaClO
NaNO2
NH4Cl
NH4F
NH4CN
C5H5NH+CN-

Table 2: Data for Determining the Ka of an Acidic Cation or Kb of a Basic Anion

Work: To show your work you can either hand write it, take a picture and insert here. Make sure it is readable. Or you can type by filling in the table below. And type remaining work below. To up your ICE table insert a table that has 8 columns and 4 rows. The first row will contain your equation for the ION that underwent Hydrolysis. Leave the first cell blank, in the second cell enter your ion, 3^rd cell your +, 4^th cell your water molecule, 5^th cell your arrow, 6^th cell your 1^st product, 7^th cell your +, 8^th cell your 2^nd product. Label rows 2-4 I, C and E. Fill in your numbers in the appropriate locations in the ICE table

NaCN Dissociation Equation:
Na+ Hydrolysis Equation:
CN- Hydrolysis Equation:
Calculation for [H3O+]:
Calculation for [OH-]:

Questions: Answer the following questions

1. For salts 7, 8, and 9 you would not have been able to predict acidic, basic or neutral from the salt formula alone. Why is that the case?

2. Go back to the simulation and select the ion pair for NaF. On the concentration bar, adjust the concentration. What do you notice happens to the pH as you increase the concentration? Provide an explanation for why that is the case.

3. Go back to the simulation and select the ion pair for NH₄Cl. On the concentration bar, adjust the concentration. What do you notice happens to the pH as you increase the concentration? Provide an explanation for why that is the case.

4. How should the pH of a 0.1 M solution of NaC₂H₃O₂ compare with that of a 0.1 M solution of KC₂H₃O₂? Explain.

Answer 1

Dear Student,

Dear Student,

	0.01M salt solution	Neutral/ acidic/ basic	pH	[H3O+]	[OH-]	Ka or kb
1	NaCl	Neutral	7	1*10-7	1*10-7
2	NaF	Basic	7.57	2.69*10-8	3.71*10-7
3	NaCN	Basic	10.70	1.99*10-11	0.0005
4	NaClO	Basic	9.73	1.86*10-10	5.37*10-5
5	NaNO2	Basic	7.67	2.13*10-8	4.67*10-7
6	NH4Cl	Slightly acidic	5.63	2.34*10-6	4.26*10-9
7	NH4F	Acidic	6.21	6.16*10-7	1.62*10-8
8	NH4CN	basic	9.32	4.78*10-10	2.08*10-5
9	C5H5NH+CN-	Slightly basic	7.29	5.12*10-8	1.94*10-7

Part 2:

Sodium cyanide is the conjugate base of hydrocyanic acid, HCN. As a consequence it goes through a hydrolysis reaction when added to water.

Therefore, w the reactione have to consider

CN^-   +H2O⟷HCN+OH⁻

	CN-	+	H2O	⇄	HCN	+	OH-
Initial Concentration)I)	0.1M		-		--		--
C	-x		-		+x		+x
E	0.1-x				xM		xM

Part 2:

Calculating [H₃O⁺]:

pH = - log [H₃O⁺].

[H₃O⁺] = 10^-pH  

Here pH=10.70

[H₃O⁺] = 10^-10.70

[H₃O⁺] =1.99*10^-11M

Calculating [OH^-]:

pH + pOH = 14

pOH = 14- pH

pOH = 14- 10.70

pOH = 3.3

pOH = - log [OH^-]

[OH^-] = 10^-3.3

[OH^-]=0.0005M

Kb= [HCN][OH−]/[CN−] =

From the table:

[x]=[OH^-]=[HCN]=0.0005(calculated above)

Kb= (0.0005*0.0005)/(0.01-0.0005)

Kb=0.000026

Kb=2.6*10^-5

NaCN Dissociatio equation: NaCN⟷ Na⁺ + CN^-

Na+ hydrolysis equation: NaCN+H2O⟷CN⁻ + Na⁺

CN- hydrolusis equation: CN^-   +H2O⟷HCN+OH⁻

****When the concentration of NaF was increased from 0.01 to 0.1 pH increases from 7.57 to 8.07 8. This is because, pH depends on concentration of H+ ions as it is negative of log of H+ ions’ concentration. As concentration is intensive property, pH is intensive too

*****When concentration of NH4Cl increases from 0.01M to 0.1M pH value decreases, means from 5.63 to 5.13. this is because

Step 1: ammonium chloride dissociation

NH4Cl -> NH4+ + Cl-

Step 2: ammonium acts as a weak acid

NH4+ + H2O <-> H3O+ + NH3

So as concentration increases [H3O+] increases acidity increase, pH value decreases

****CH3 COO Na is a salt made when NaOH and CH3 COOH react with each other.

CH3 COO K is a salt made when KOH and CH3 COOH react with each other. in both cases the alkali is strong and the acid is the same weak acid.

when these salts are placed in water, the conjugate base from the weak acid (CH3 COO-) is involved in hydrolysis reaction. the K+ and Na+ produced do not react with water. hence the final pH of the solution will be the same in the two salts.

Thanks & Brest wishes