Question

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The following table, used for question 16, is partially complete: (ОН) pH 2.68 pОН ener 8.4 x 10 euen 16. (3 points) Fill in the six empty cells in the table. The following table, used for questions 2-8, is partially complete: Ka 1.8 x 10 pka Conj. Base formula pKb Acid Formula CH3COOH HOBI HNO2 5.40 3.14 2.2 x 10 Sec 17. (6 points) Fill in the twelve empty cells of the table above. 18. What is the strongest acid in the first column of the table? 19. What acid, and what conjugate base would be most suitable for making a pH = 3.0 buffer? acid conjugate base 20. Write the complete chemical reaction for the acid dissociation of HNO2: 21. KOH is a strong base. What is the pH of 0.040 M KOH solution? 22. (3 points) What is the pH of 25.0 mL of 0.20 M KOH mixed with 5.0 mL of 1.1 M HCI (assuming the final volume is 30.0 mL)? 23. Glycine becomes a conjugate base at high pH and has the following structure: H2NCH2C00". When acid is added, the amino group's N is a stronger base than the carboxylate group's O. Write the formula of the zwitterion that would result when acid is added to glycine's anion: 24. Given the name "diamminediaquanickel(II)ion", write its formula:

Answer 1

16.

[H^+]	[OH^-]	pH	pOH
2.09 x 10^-3 M	4.77 x 10^-12 M	2.68	11.32
8.4 x 10^-9 M	1.19 x 10^-6 M	8.08	5.92

Note :

* pH = - Log[H^+]

* pOH = - Log[OH^-]

* [H+] = 10^-pH

* [OH-] = 10^-pOH

* pH + pOH = 14

17.

Acid formula	Ka	pKa	conj.base formuls	pKb
CH3COOH	1.8 x 10^-5	4.74	CH3COO^-	9.26
HOBr	2.51 x 10^-9	8.60	OBr^-	5.40
HNO2	7.24 x 10^-4	3.14	NO2^-	10.86
HSeO4^-	2.2 x 10^-2	1.66	SeO4^2-	12.34

Note :

* pKa = - LogKa

* pKb = - LogKb

* pKa + pKb = 14

* Conjugate base has one proton less than its conjugate acid.

18.

Higher is the Ka higher is acidic nature.

Since HSeO4^- has higher Ka and hence it is stronger acid.

19.

HNO2 and NO2^- form best buffer of pH = 3.0 because the pKa HNO2 is 3.14 which is nearer to pH of buffer.

Acid : HNO2

Conjugate base : NO2^-

20.

HNO2 (aq) + H2O (l) $\rightleftharpoons$ H3O^+ (aq) + NO2^- (aq)

21.

[OH-] = [KOH] = 0.040 M

pOH = - Log[OH-] = - Log(0.040) = 1.40

pH = 14 - pOH = 14 - 1.40 12.6

22.

MaVa = 1.1 x 5.0 = 5.5

MbVb = 0.20 x 25.0 = 5.0

MaVa - MbVb = 5.5 - 5.0 = 0.5

Volume of solution = 30.0 mL

Since, MaVa > MbVb , the resulting solution has excess H^+ ions.

[H^+] = [ MaVa - MbVb ] / Volume of solution = 0.5 / 30.0 = 0.0167 M

pH = - Log[H^+] = - Log(0.0167 ) = 1.78

23.

+^H3N - CH2 - COO^-

Zwitter ion : N carries positive sign adn O carries negative sign.

24.

[Ni(H2O)2(NH3)2]^2+