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) The Ka for HCN is 4.9 ~ 10-10. What is the value of Kb for CN-? 2) Using the data in the table for 0.10 M aqueous species at 25 °C, which conjugate base below is the weakest base? Explain your answer for full credit! Acid ка. НОАс 1.8 x 10-5 HC7H502 6.3 10-5 HNO2 4.5 x 10-4 HF 6.8 x 10-4 3) The Ka for HCN is 4.9 10-10. What is the pH of a 0.068 M aqueous solution of potassium cyanide?

Thank you!

Answer 1

Given that; ka for HON IS 409 x 10-10 Ht+ CN Ka= 4•9 x 1010 HCN (ada) (conjugate Bose) ka = [ht] [CN) --M) EHCN Now, CN + H₂O HCN tour CCB) Kb= ? Kb = [HCN) toul (2) [CNJ (H₂O) multiply both the tqn Ka x Kb = ther) [tat)y then[OUT ENEN CCNJ (H₂O) Kax Kb= Kw 1oox 10-14 Now, kb= loox 1014 40g & 1010 Kb = (2) hence lessa HA H+ + A- (Aced) Ссв) Strongen er the aud (more will be ka) will be its strength of So weakest the conjugate base. conjugate base (weak base) larger the ka

Order of Ka Kas for 1..8x 10-5 HOA HCA HISO, HNG, HF are < 613 x 10 5 < 4.5 X 10-14 < 6.8 10-4 CHNO) (RE) CHOAC) (HGHSB) Hence conjugate base strength order dore F- No C4 H50 < OAC" so HF will be ate base. the weakest conjugate produces (3) HEN Ka= 409 x 1010 H+ + CN- (18) ka x kg = 10-14 kb= 10-14 2.040 x 10-5 uod & 10-10 CN req) + H₂0 (1) HCN ton- Kb= [HCN COHO [HCM (DH) 20040 x 10-5 CONJCH,0) ta) ~ Neguet CN HCN OH- (2)(x) 2 = 2.040x10-5 0.068M 16.068-23 UADOR geoler 3.628 X10 4 0 D 0. 068-2 pon--log (2) = -log(80628 x 10") 3044 PH= 14-pou= I= 14- 2044 ph= 10.5596