CHEM 6B Lecture Notes - Lecture 10: London Dispersion Force, Bravais Lattice, Quadratic Formula

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burgundyporpoise729

6 Apr 2018

School

UCSD

Department

Chemistry and Biochemistry

Course

CHEM 6B

Professor

Sandrine Bernoilles

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Document Summary

Problem, phosphoric acid, h3po4 is a triprotic acid. The pka"s are as follows: pka1 = 2. 16, pka2 = 7. 21, pka3 = 12. 32, calculate the [ ] of h3po4, h2po4-, hpo4 2-, and po4 3- and ph @ eq. 0. 100m solution: h3po4 + h2o h2po4- + h3o+ 0. 100 - x x x: 6. 9 x 10^-3 = x^2 / 0. 100 - x > 5% cannot approximate, quadratic formula: x = 0. 023. [h3o+] = 0. 023: h2po4- + h2o h3o+ + hpo4- [if h3o+ > 10^-7, include in. 0. 023 + x x: 6. 2 x 10^-8 = (0. 023 +x)(x) / (0. 023 - x, neglect +x and -x, check 5% If the cation is derived from a strong base, it does not affect the ph. But if the cation is derived from a weak base, it makes the solution acidic. If the anion is derived from a strong acid, it does not affect the ph.

Related Questions

Salts formed by reaction of a strong acid with a weak base areacidic, and salts formed by reaction of a weak acid with a strongbase are basic. This is due to the equilibrium the conjugate haswith its acid or base. For example, sodium bicaronate NaHCO3contains the conjugate of the weak acid H2CO3. The conjugate willreform the acid through the equilibrium

HCO3?(aq)+H2O(aq)?H2CO3(aq)+HO?(aq)

and, thus, acts as a weak base. A solution of NaHCO3yields abasic solution.

If a salt is derived from a weak acid and a weak base, the pH ofthe solution depends on the relative acid strength (Ka) ofthe cation and base strength (Kb) of the anion. IfKa>Kb, the solution of the salt is acidic. IfKa<Kb, it is basic. If Ka?Kb, it is neutral.

Given that

Ka for HCN is 4.9

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

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

CHEM 6B Lecture Notes - Lecture 10: London Dispersion Force, Bravais Lattice, Quadratic Formula

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