Question

Hey guys, I am having a lot of trouble with my homework. Could you please help? I thumbs up anyone who gives a thorough explanation and shows the work! I really need to understand these concepts!

1) At 25 °C, the pH of a 0.15 M aqueous solution of CsZ (the cesium salt of HZ) is 10.70. What is the Ka for HZ? 2) Calculate the pH of 0.586 M anilinium hydrochloride (C6H5NH3CI) solution in water at 25 °C, given that Kb for aniline is 3.83 x 10-4. 3) Kb for NH3 is 1.8 x 10-5. What is the pH of a 0.45 M aqueous solution of NH4Cl at 25.0 °C?

Thanks in advance!

Answer 1

Am - The pH of weak acid and strong base is given by - pr 7 - 1 pka - I lege Given - PH? values and solve for pka C= o. ISM 10.70. put the hewe, 10.707-1 fka - lag 0.15

10.70 = 7 - 1 pka to. 412. ex (10.70) = 14-pka +0.824 pla -6.57b. we know that pka Ka = 10 log ka. -pkai hence Kaz 10- (-6.576) Ka= 2.65 x 10 7

Am-2 Х According to reactien ean:- C6Hg NH3 + H2O Initial - (6H, NH2 + H30" 0.586 Change Elu 10.586-x And ka= [caths NH2) [m2ot] Corary [C6HI NH₂ t]. We know that Ka = Kw kb. kw = 1 x 10th 23.83x10-4 6 Kb hence 2.68 10-11 Ka= 1810-14 3.83X10-4 put all the values and selve for & 2 2. 6x 10ll [x] [x] [0.586-x

@ or 2.64 101 = x² 0.586 ka value is O vey dmall . x = 3.87810-6 When then, ph= -log [H301] = x = 3.87X106 [ dot]. pn = -log 73.8 78100) po 14 = 5.4 @

Am-3 WHY + H2O - NH₃ + H₂Ot. I С. + x + x E Cour-x) x And ka- (NH₃] [naot] Ently t]. we know that kaa kw Kb -10 =5.5 x 70 Kaz 1x10th 1.8x105

put all the values and solu olving for x. ka= [n] [u Coous-x). ka x2 o.us. 5.5x10 10 = x2 N 0.45 ka value is very small) 2.48 10 10 = x² 2 x 1.55810T when [ol3ot) = = X = 1.55 X 10.5. then pH = -log {4130t pH = -log (1.5x105] apu H = 4.8