Question

Titration Curves of Strong and Weak Acids and Bases. Pre-lab questions. Sed and weak acids ng acids cont 214-2343 -110.51 Explain the difference between a strong acid and a weak acid. A Strong Acd die Insed and weak are ionsrzed partially. Moreover Strong oud a higher concentration of Hydrogen 2. Calculate the pH of an HCl solution in which the (H30'] = 1.75 x 10 M. DH- og IH30-] [H2O PH- - 1.75% 103] PH=72.75 3. Calculate the pH of a NaOH solution in which the (OH) = 3.7 x 10 M. POH = -log [OH- ] Eiven LOH- ] 3.7x10. POH= -log T3.7x104)3.43 Now, PH + POH=19.PH = 14-3.43 4. Calculate the [H,O*] when pH = 7.42 -1973 PH = -109 [H307] 7.42= -log LH30 - [H30] = 10-7.42 [H3O+] =3,8810-8 egarth thum 10 base 5. Calculate the [H30*] when pOH = 5.43 PH= 14 - Pot=14-5.43 8.57 [H3O+] =26 Tog-TH30 =0.57 LH 30+)- 10-8.57 Calculate the pH of 1.00 M acetic acid, HCHO2 (aq). (Ka = 1.8 x 10%) - 10 CH3COCH=CH-Coot 0 -X + X CHCOOH 11.0-x) + lculate the Ka of a 0.50 M HCHO2 with a pH of 2.56. 3C1+2C42-600] TH+] PH = 2.36 X2=1.8x10 - O kaICH3COOTH+ +X CH200 14 = 2.36 Trr.8x10-S X. Eilen ka-1180 16- +3 CH2 - CH2 - Cool x Ka= x² x²=0.5xka 14+ ax=4.74 +30+1=-05 xfoSxkajo s te-log [40. S] glx]-log To stron 12 **=4.2x10-3

Answer 1

Solution Lob kolor *C( Jully lonised but weak auids Ak One Pntyionisedin Solution A the Same Con Cen tnation Strong aeids have a waher Cencentnation of drogen ions then usak acids o e Hrochlonie ai is a reng auid and ace tic auid useak aud Given, CH30 B0for0std) 175 X10-3 (M). wailantas -: Ocff 11 2.757 2.757 am Hel solution in ohich tu CH3ot) = .X103 (m) o od b>oab The PH-

Grdven CoH- 3 7I- (M) ye 3 -Lg (3. x 10-4)3.432 poH-- CoH PH= 14-3 432-t0.56g 2.x / The pH- 10. S68 of a Naot Solution in oich the Co Given pH7.42 -7.42 3.8x 10-8 H30 10- PH 10 H3ot3.8x10-8 (m) aid Con centra hon Ceel2 C Ced Given pon -s43 5 . pa.thcalag om :: PH 14-PoH - 14-5.43 Q.57 Xx CHgotJ107 8:5 .P-Lo CH30t) 2.692x10-9 : Con Centralin f CH30 2C92x 10- (m) CH3 Curo H C1-2) CHaCro+ Ht degnee of dissaialien) DHT) EH2) ka = [CH3LeTH 2 KA nt Ka n- Ka o Ka- .Ka r Given: Ka - 1.8X10-(M) f X P3 Td ! 2.1 hup 0 g

(8x0-52 +4 AI8X- 8 1D-S 4.234x10-3 HA Tte Concnahon of t4:204 X10-3 (m) . . H-l23y K-3) : P .373 + 6.5(1-2) 0.5 ofmee of dissocia tion ka 0522 0.5 x2+ Ka Ka= O Guiven, pH Q.55 H 256Le (o5)i) =-09 CHt) .75y ie-3 ) 0.5K O 5 x 10- 2.56 (n=5.508 X.10-3) Pu hng He vaut of in esua hon- 5.508K10-3 Ka -ka 0 05x (5.508 A103)2 151 69X 10-5 0.9945 Ka 1 2 1 78-1 =The Ka value of ty Ho2 aeid unl be 1.5210-. Ka t 535X 10-5