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pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

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Page 1: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

pOH = -log [OH-]

[H+][OH-] = Kw = 1.0 x 10-14

-log [H+] – log [OH-] = 14.00

pH + pOH = 14.00

Page 2: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

Page 3: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

تدريب : يساوي [ +H] لمحلول يحتوي على تركيزpOH و pHاحسب قيمة )1(

2.0 x 10-3 .موالر 3.70 تساوي pH في محلول قيمة [-HO] و [+H]احسب )2( جم 0.31 مل ويحتوي على 400 لمحلول حجمه pOH و pHاحسب )3(

Ba(OH)2من هيدروكسيد الباريوم

الفصل الثاني بالباب وحلول مسائل كتاب في اإلجابة توجد .7للتأكد

Page 4: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)H2O

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

Page 5: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)H2O

KOH (s) K+ (aq) + OH- (aq)H2O

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)H2O

F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Bases are weak electrolytes

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Page 6: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

HA (aq) H+ (aq) + A- (aq)

Ka =[H+][A-][HA]

Ka is the acid ionization constant

Page 7: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M

Page 8: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

x2 = 6.95 x 10-5

x = 0.0083 M

pH = -log[H+] = -log[0.0083] = 2.08

Page 9: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0081 x = - 0.0081

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

وهي : مباشرة التالية المعادلة تطبيق يتم آخر حل

= 5.7 x 10-4 x 0.122 = 0.0083 M aa C . K ] H[

Page 10: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

aa C . K ] H[

(1) Find the pH of a 0.135 M aqueous solution of periodic acid (HIO4), for which Ka = 2.3 10-2. 

A. 1.25 B. 3.28 C. 1.17D. 1.34 E. 1.64

 

(2) Find the pH of a 0.183 M aqueous solution of hypobromous acid (HOBr), for which Ka = 2.06 10-9. 

A. 4.71 B. 8.69 C. 3.97D. 4.34 E. 9.28

= 2.3 x 10-2 x 0.135 = 0.0557 M

pH = - log [0.0557] = 1.25

= 2.06 x 10-9 x 0.183 = 1.94 x 10-5 M

pH = - log [1.94 x 10-5] = 4.71

aa C . K ] H[

Page 11: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Page 12: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

مل 200 في محلول حجمه +Ba2 و -(OH) احسب التركيز بالموالرية لكل من أيونات مثال : . Ba(OH)2 جم من هيدروكسيد الباريوم 0.612ويحتوي على

جم 0.612 يتم أوال حساب عدد موالت هيدروكسيد الباريوم بداللة كتلته الحـل : جم/مول 137.34 + )16 + 1 ( 2 = 171.34 = .(Mol. wt)ووزنه الجزيئي

13

mol.gm 34.171

gm 612.0 10 6.3

= عدد موالت هيدروكسيد الباريوم = مول إذا

)L(V

)mole(n

1000

200106.3

0179.0M3

والتركيز بالموالرية M =

قوية +Ba(OH)2(S) Ba2 قاعدة(aq) + 2 HO-1

(aq)

= : بعد معرفة التركيز الحجمي بالموالرية . يتم كتابة معادلة تأين اإللكتروليت Ba(OH)2ثانيا

+Ba2 و -(OH)الموزونة وذلك لحساب تركيز كل من أيونات

موالر 0.0179 صفر موالر صفر موالرفي البداية: موالرx 0.0179 + - 0.0179 2+ 0.0179بعد الذوبانية:

صفر موالر موالر0.179 موالر 0.0358النتيجة:

موالر . 1 = 2 × 0.0179 = 0.0358-(HO)2تركيز أيونات الهيدروكسيد للمحلول. pHواآلن أحسب قيمة الـ موالرBa2+ = 0.179وتركيز أيونات الباريوم

pH = 12.55اإلجابة :

Page 13: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

الـ قيمة قيمة xتهمل Kbالنمن 1اصغر

What is the pH of a 0.40 M Ammonia solution? (if the Kb = 1.8 x 10-5)

(a) 11.43 (b) 12.43 (c) 2.57 (d) 3.57

NH3 + H2O NH4+ + HO-

0.4M 1M 0 0

0.4 – x 1M x x

Kb = [x][x] ÷ [0.4-x] = 1.8 x 10-5

X2 = 1.8 x 10-5 X 0.4 = 7.2 X 10-6 X = 7.2 X 10-6 = 2.68 X 10-3

pOH = - log 2.68 x 10-3 = 2.57 pH = 14-2.57 = 11.43

وهي : مباشرة التالية المعادلة تطبيق يتم آخر حل

= 1.8 x 10-5 X 0.4 = 2.7 x 10-3

pH = -log 2.7 x 10-3 = 2.57 pH = 14-2.57 = 11.43

bb C . K ]HO[

Page 14: POH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00

bb C . K ]HO[ يتم التطبيق المباشر على

Kb = 1.8 x 10-10حيث المعطاة بالمسألة هي : ثابت االتزان للقاعدة الضعيفة األمونيا موالر Cb = 0.10وتركيزها

M 10 342.1 1.0 10 8.1 ]HO[ -310

تأينه 0.10محلول مثال : ثابت األمونيا من . Kb = 1.8 x 10-5موالرقيم للمحلول .pOHو pHاحسب

pOH = - log 1.342 x 10-3 = 2.87

pH = 14 – 2.87 = 11.13