CHEM 101 Lecture Notes - Lecture 9: Coordinate Covalent Bond, Valence Electron, Ammonium Sulfate

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amethystserval28

24 Sep 2016

School

Binghamton University

Department

Chemistry

Course

CHEM 101

Professor

Steven Boyer

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Document Summary

Ion formed from a group of atoms (held together by covalent bonds) through loss or gain of electrons: determined in the same way as those for ionic compounds that contain monatomic ions o o. The positive and negative charges present must add up to zero. Required polyatomic ions: nitrate, nitrite, ammonium, sulfate (hydrogen sulfate/ bisulfate, sulfite (hydrogen sulfite/ bisulfite, phosphate (hydrogen phosphate, phosphite (hydrogen phosphite, carbonate (hydrogen carbonate/ bicarbonate, acetate, cyanide, hydronium, hydroxide. Covalent bond: chemical bond formed through the sharing of one or more pairs of electrons between two atoms: molecular compound (covalent compound): a compound in which atoms are joined through covalent bonds. Ionic bonds form between a metal and nonmetal (also poly atomic ions) All ionic compounds are solids at room temperature. Soluble ionic solids form aqueous solutions that conduct electricity. Covalent compounds have a molecule as its basic structural unit.

Related Questions

Below is a list of some common polyatomic ions for your reference.

Ion Name	Ion Formula	Ion Name	Ion Formula
carbonate	$C O_{3}^{2 -}$	phosphite	$P O_{3}^{3 -}$
nitrite	$N O_{2}^{-}$	phosphate	$P O_{4}^{3 -}$
nitrate	$N O_{3}^{-}$	sulfate	$S O_{4}^{2 -}$
ammonium	$N H_{4}^{+}$	chlorate	$C l O_{3}^{-}$
hydroxide	$O H^{-}$	perchlorate	$C l O_{4}^{-}$

Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. A crystalized product is a solid. The other product, lithium chlorate, remains dissolved in solution. Consider the other product and its phase, and then write the balanced molecular equation for this precipitation reaction.

Express your answer as a chemical equation including phases. Type an underscore (_) or a carat (^) to add subscripts and superscripts more quickly.

tealdog287

Calculate the pH change that results when 12 mL of 5.9 M NaOH is added to 763 mL of each the following solutions. (See the appendix.)

(a) pure water

(b) 0.10 M NH₄Cl

(c) 0.10 M NH₃

(d) a solution that is 0.10 M in each NH₄⁺ and NH₃

Acid Name	ConjugateAcid	K_a	pK_a	Conjugate Base	Base Name
perchloric acid	HClO₄	>>1	< 0	ClO₄^1-	perchlorate ion
hydrohalic acid	HX (X=I,Br,Cl)	>>1	< 0	X^1-	halide ion
sulfuric acid	H₂SO₄	>>1	< 0	HSO₄^1-	hydrogen sulfate ion
nitric acid	HNO₃	>>1	< 0	NO₃^1-	nitrate ion

hydronium ion	H₃O¹⁺	1.0	0.00	H₂O	water
iodic acid	HIO₃	0.17	0.77	IO₃^1-	iodate ion
oxalic acid	H₂C₂O₄	5.9 x 10^-2	1.23	HC₂O₄^1-	hydrogen oxalate ion
sulfurous acid	H₂SO₃	1.5 x 10^-2	1.82	HSO₃^1-	hydrogen sulfite ion
hydrogen sulfate ion	HSO₄^1-	1.2 x 10^-2	1.92	SO₄^2-	sulfate ion
phosphoric acid	H₃PO₄	7.5 x 10^-3	2.12	H₂PO₄^1-	dihydrogen phosphate ion
hydrofluoric acid	HF	7.2 x 10^-4	3.14	F^1-	fluoride ion
nitrous acid	HNO₂	4.0 x 10^-4	3.40	NO₂^1-	nitrite ion
lactic acid	HC₃H₅O₃	6.4 x 10^-5	3.85	C₃H₅O₃^1-	lactate ion
formic acid	HCHO₂	1.8 x 10^-4	3.74	CHO₂^1-	formate ion
hydrogen oxalate ion	HC₂O₄^1-	6.4 x 10^-5	4.19	C₂O₄^2-	oxalate ion
hydrazoic acid	HN₃	1.9 x 10^-5	4.72	N₃^1-	azide ion
acetic acid	HC₂H₃O₂	1.8 x 10^-5	4.74	C₂H₃O₂^1-	acetate ion
carbonic acid	H₂CO₃	4.3 x 10^-7	6.37	HCO₃^1-	hydrogen carbonate ion
hydrogen sulfite ion	HSO₃^1-	1.0 x 10^-7	7.00	SO₃^2-	sulfite ion
hydrosulfuric acid	H₂S	1.0 x 10^-7	7.00	HS^1-	hydrogen sulfide ion
dihydrogen phosphate ion	H₂PO₄^1-	6.2 x 10^-8	7.21	HPO₄^2-	hydrogen phosphate ion
hypochlorous acid	HClO	3.5 x 10^-8	7.46	ClO^1-	hypochlorite ion
ammonium ion	NH₄¹⁺	5.6 x 10^-10	9.25	NH₃	ammonia
hydrocyanic acid	HCN	4.0 x 10^-10	9.40	CN^1-	cyanide ion
hydrogen carbonate ion	HCO₃^1-	4.7 x 10^-11	10.33	CO₃^2-	carbonate ion
hydrogen phosphate ion	HPO₄^2-	4.8 x 10^-13	12.32	PO₄^3-	phosphate ion
hydrogen sulfide ion	HS^1-	1.3 x 10^-13	12.89	S^2-	sulfide ion
water	H₂O	1.0 x 10^-14	14.00	OH^1-	hydroxide ion

ammonia	NH₃	<<10^-14		NH₂^1-	amide ion
hydroxide ion	OH^1-	<<10^-14		O^2-	oxide ion

What is the pH of a solution made by dissolving 4.3 g of K₂CO₃ and 1.6 g of KHCO₃ in 500. mL of water? (See the appendix.)

Table 3 An Acid-base table

Acid Name	ConjugateAcid	K_a	pK_a	Conjugate Base	Base Name
perchloric acid	HClO₄	>>1	< 0	ClO₄^1-	perchlorate ion
hydrohalic acid	HX (X=I,Br,Cl)	>>1	< 0	X^1-	halide ion
sulfuric acid	H₂SO₄	>>1	< 0	HSO₄^1-	hydrogen sulfate ion
nitric acid	HNO₃	>>1	< 0	NO₃^1-	nitrate ion

hydronium ion	H₃O¹⁺	1.0	0.00	H₂O	water
iodic acid	HIO₃	0.17	0.77	IO₃^1-	iodate ion
oxalic acid	H₂C₂O₄	5.9 x 10^-2	1.23	HC₂O₄^1-	hydrogen oxalate ion
sulfurous acid	H₂SO₃	1.5 x 10^-2	1.82	HSO₃^1-	hydrogen sulfite ion
hydrogen sulfate ion	HSO₄^1-	1.2 x 10^-2	1.92	SO₄^2-	sulfate ion
phosphoric acid	H₃PO₄	7.5 x 10^-3	2.12	H₂PO₄^1-	dihydrogen phosphate ion
hydrofluoric acid	HF	7.2 x 10^-4	3.14	F^1-	fluoride ion
nitrous acid	HNO₂	4.0 x 10^-4	3.40	NO₂^1-	nitrite ion
lactic acid	HC₃H₅O₃	6.4 x 10^-5	3.85	C₃H₅O₃^1-	lactate ion
formic acid	HCHO₂	1.8 x 10^-4	3.74	CHO₂^1-	formate ion
hydrogen oxalate ion	HC₂O₄^1-	6.4 x 10^-5	4.19	C₂O₄^2-	oxalate ion
hydrazoic acid	HN₃	1.9 x 10^-5	4.72	N₃^1-	azide ion
acetic acid	HC₂H₃O₂	1.8 x 10^-5	4.74	C₂H₃O₂^1-	acetate ion
carbonic acid	H₂CO₃	4.3 x 10^-7	6.37	HCO₃^1-	hydrogen carbonate ion
hydrogen sulfite ion	HSO₃^1-	1.0 x 10^-7	7.00	SO₃^2-	sulfite ion
hydrosulfuric acid	H₂S	1.0 x 10^-7	7.00	HS^1-	hydrogen sulfide ion
dihydrogen phosphate ion	H₂PO₄^1-	6.2 x 10^-8	7.21	HPO₄^2-	hydrogen phosphate ion
hypochlorous acid	HClO	3.5 x 10^-8	7.46	ClO^1-	hypochlorite ion
ammonium ion	NH₄¹⁺	5.6 x 10^-10	9.25	NH₃	ammonia
hydrocyanic acid	HCN	4.0 x 10^-10	9.40	CN^1-	cyanide ion
hydrogen carbonate ion	HCO₃^1-	4.7 x 10^-11	10.33	CO₃^2-	carbonate ion
hydrogen phosphate ion	HPO₄^2-	4.8 x 10^-13	12.32	PO₄^3-	phosphate ion
hydrogen sulfide ion	HS^1-	1.3 x 10^-13	12.89	S^2-	sulfide ion
water	H₂O	1.0 x 10^-14	14.00	OH^1-	hydroxide ion

ammonia	NH₃	<<10^-14		NH₂^1-	amide ion
hydroxide ion	OH^1-	<<10^-14		O^2-	oxide ion

CHEM 101 Lecture Notes - Lecture 9: Coordinate Covalent Bond, Valence Electron, Ammonium Sulfate

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