Question

Report 1.

Make a table to show and compare the measured pH using the pH meter and the paper for each solution to the calculated pH.

#	Solution	pH (paper)	pH (pH meter)	pH (Calculated)
1	0.1 M HCl	1.0	1.115
2	0.01 M HCl	4.0	2.925
3	0.001 M HCl	3.0	2.942
4	0.1 M NaCl	5.0	5.809
5	0.1 M HC2H3O2	3.0	2.932
6	0.1 M NaOH	12.0	12.793
7	deionized water	6.0	6.664
8	Tap water	7.0	7.722
9	0.1 M NH3	9.0	10.376
10	0.1 M NH4Cl	6.0	6.275
11	0.1 M (NH4)2C2O4	8.0	7.954
12	0.1 M NaC2H3O2	7.0	7.965
13	0.1 M NaH2PO4	4.0	4.421
14	0.1 M Na2HPO4	9.0	8.952
15	0.1 M Na3PO4	12.0	12.259
16	0.1 M H3PO4	1.0	1.714

  

1. Calculate the pH for each of the above solutions.    

Report

       Prepare a table with volume of NaOH or HCl added and the resulting pH of the water and the buffer solution.  

Water				Buffer solution
Beaker 1		Beaker 2		Beaker 3		Beaker 4
Total Volume of 1 M NaOH added	pH	Total Volume of 1 M HCl added	pH	Total Volume of 1 M NaOH added	pH	Total Volume of 1 M HCl added	pH
0 mL	4.0	0 mL	4.0	0 mL	4.740	0 mL	4.674
1 drop	8.0	1 drop	3.509	1 mL	4.714	1 mL	4.527
2 drops	7.0	2 drops	3.371	2 mL	4.715	2 mL	4.302
6 drops	9.0	6 drops	2.923	6 mL	4.735	6 mL	3.806
10 drops	9.0	10 drops	2.694	10 mL	4.764	10 mL	2.585
20 drops	8.0	20 drops	2.406	20 mL	4.886	20 mL	0.979
2 mL = (40 drops)	11.0	2 mL	2.139	30 mL	5.073	30 mL	0.819

1. Calculate the concentration of the sodium acetate and acetic acid buffer solution,
2. Calculate the expected pH of the buffer solution. Compare this to the measured pH

1. Create a table with the mass of KHP and volume of NaOH solution used in each titration. Show the moles of KHP, moles of NaOH and molarity of NaOH.

2. Show how to calculate the moles of KHP, moles of NaOH and molarity of the NaOH solution using your values. . Use at least 6 trials or the complete class data to determine the average and standard deviation of the molarity.

Answer the following question:  

1. If we know the mass of NaOH used to make the solution, why did we use the primary standard to determine the concentration of the NaOH solution?

Answer 1

two questions/Reports , with so many sub-parts (calculation ) as per guideline only 1st report question will be answered (completely ) :

pH calculation for each solution :

1. 0.1 M HCl

pH = - (log[H⁺ ]) = - (log[0.1 ]) = 1.0

2. 0.01 M HCl

pH = - (log[H⁺ ]) = - (log[0.01 ]) = 2.0'

3. 0.001 M HCl

pH = - (log[H⁺ ]) = - (log[0.001 ]) = 3.0

4. 0.1 M NaCl (it is salt of strong base and strong acid , thus solution will be neutral , as salt hydrolysis will give equal amount of H⁺ and OH^- )

pH = 7.0

5. 0.1 M CH₃COOH (weak acid ) ka = 1.8*10^-5

for weak acid [H⁺ ] = (ka*(HA))^1/2 = (1.8*10^-5*(0.1))^1/2 = 1.34*10^-3 M

pH = - (log[H⁺ ]) = 2.87

6. 0.1 M NaOH

[OH^- ] = 0.1 M

pH = 14 -pOH = 14 - (-log[0.1 ] = 13

7. De-ionized water kw =  [H⁺ ][OH^- ] = 10^-14

[H⁺ ]= 10^-7

pH = - (log[H⁺ ]) =7.00

8. tap water may contain impurities, which may make it acidic or basic : so, it cannot be calculated without any other reference.

9. 0.1 M NH₃ (weak base ) kb = 1.8*10^-5

[OH- ] = (kb*(B))^1/2 = (1.8*10^-5*(0.1))^1/2 = 1.34*10^-3 M

pH = 14 -pOH = 14 - (-log[1.34*10^-3 ] = 11.13

10. 0.1 M NH₄Cl (salt of weak base and strong acid : solution will be acidic)

NH₄⁺ (aq)+ 2H₂O (l) <===>NH₄OH (aq)+ H₃O⁺ (aq)

kb = 1.8*10^-5

[H⁺ ]= ((kh (NH₄Cl ))^1/2 = ((kw/kb* (NH₄Cl ))^1/2 = ((5.6*10^-10 (0.1 ))^1/2 = 7.48*10^-6

pH = - (log[H⁺ ]) =5.13

11. 0.1 M (NH₄)₂C₂O₄ (salt of weak base and weak acid : pH will depend on Ka and Kb values : for oxalic acid we will consider ka2 )

kb = 1.8*10^-5 ; Ka2 = 1.5*10^-4

kh = (kw / kb*ka2) = 3.7*10^-6

[H⁺ ]= (kh)^1/2 ka2 = (3.7*10^-6)^1/2*1.5*10^-4 = 2.89*10^-7

pH = - (log[H⁺ ]) = 6.54

12. 0.1 M  CH₃COONa (Salt of weak acid and strong base : solution will be basic)

CH₃COO- (aq)+H₂O (l) <===>CH₃COOH (aq)+ OH- (aq)

ka = 1.8*10^-5

[H⁺ ]= = ((kw* ka /(CH₃COONa ))^1/2 = (( 1.8*10^-19 /(0.1 ))^1/2 =1.34*10^-9

pH = - (log[H⁺ ]) =8.87

13. 0. 1 M NaH₂PO₄  

(Salt of weak cojugated cation and amphoteric anion : k1 (7.5*10^-3) ,k2 (6.2*10^-8),k3 (10^-12)for H₃PO₄ will be used )

pH = 1/2 (pk1 + pk2) = 4.66

14.

0. 1 M Na₂HPO₄

(Salt of weak cojugated cation and amphoteric anion : k1,k2 (6.2*10^-8),k3 (10^-12)for H₃PO₄ will be used )

pH = 1/2 (pk2 + pk3) = 9.6

15.

0. 1 M Na₃PO₄  

(Salt of weak acid and strong base : solution will be basic)

PO4^3-(aq)+H₂O (l) <===> HPO4^2-(aq)+ OH- (aq)

ka3 = 10^-12

[H⁺ ]= = ((kw* ka /(CH₃COONa ))^1/2 = (( 10^-26 /(0.1 ))^1/2 =3.16*10^-13

pH = - (log[H⁺ ]) =12.5

16

0. 1 M H₃PO₄  ( weak acid : ka1 = 7.5*10^-3))

for weak acid [H⁺ ] = (ka*(HA))^1/2 = (7.5*10^-3)*(0.1))^1/2 = 2.74*10^-2 M

pH = - (log[H⁺ ]) = 1.56