1.Calculate the H+ in 1X10^-(4)M HCN(aq)
2.Calculate the fractions of HCO3-(aq) at PH 10(KA1=4.3X10^-7.KA2=4.8X10^-11).
3.Calculate the PH of a 0.1M NH4CN(aq)(KA(NH4OH)=1.8X10^-5..KA(HCN)=6.2X10^-10)
4.Calculate the PH value of 50ml 0.1M HOAC(KA=1.8X10^-5)titrater with 50ml 0.1M NAOH added
5.Calculate the PH of 1.0M solution of NAH2PO4(aq) (H3PO4 KA1=7.5X10^-3.KA2=6.2X10^-8.KA3=4.8X10^-13)
6.Show that the solubility of AL(OH)3 as a function of H+.obeys the eqution: S=(H+)^KsP/KW^3+KKw/(H+) where S= solubility =(AL3+)+(AL(OH)4) and K is the equilibrium constant for AL(OH3)+4OH-(aq)=AL(OH)4-
7.In the titration of 100ml of a 0.05M solution of acid H3A(kA=10^-3.KA2=5X10^-8.KA3=2X10^-12. Calculate the volume of 1M NAOH required to reach PH value of 9.5.
8.What is the solubility of AgCl(S) in 10M NH3(KSP(agcl)=1.6x10^-10 Kf1(agnh3)=2.1x10^3 kf2(ag(nh3)2)=8.2x10^3
(非常急)大學普化題目求解
2015-01-13 6:22 am
回答 (6)
2015-01-13 10:01 pm
✔ 最佳答案
分開八題發問，每題五分，你得到部分題目答案的機會較大。2015-01-13 14:01:59 補充：
1.
HCN(aq) ⇌ H⁺(aq) + CN⁻(aq) .. Ka = 6.2 × 10⁻¹⁰
H2O(l) ⇌ H⁺(aq) + OH⁻(aq) .. Kw = 1 × 10⁻¹⁴
At eqm :
Let [CN⁻] =p M and [OH⁻] =q M
[HCN] = [(1 × 10⁻⁴) - p] M ≈ 1 × 10⁻⁴ M(Assume that (1 × 10⁻⁴) >> p)
[H⁺] = (p + q) M
Ka = [H⁺][CN⁻]/[HCN]
(p + q)q/(1 × 10⁻⁴) = 6.2 × 10⁻¹⁰
(p + q)q = 6.2 × 10⁻¹⁴ ...... [1]
Kw = [H⁺][OH⁻]
(p + q)p = 1 × 10⁻¹⁴ ...... [2]
[1]/[2] :
q/p = 6.2
q = 6.2p ...... [3]
Sub. [3] into [2] :
(p + 6.2p)p = 1 × 10⁻¹⁴
7.2p² = 1 × 10⁻¹⁴
p² = (1 × 10⁻¹⁴)/7.2
p = 3.7 × 10⁻⁸
Sub p = 3.7 × 10⁻⁸into [3] :
q = 6.2(3.7 × 10⁻⁸)
q = 2.3 × 10⁻⁷
[H⁺] at eqm = p + q = 2.7 × 10⁻⁷ M
====
4.
HOAc(aq) + OH⁻(aq)→ OAc⁻(aq) + H2O(l)
Neglecting the dissociation of HOAc, the neutralization is complete.
[HOAc]o = 0 M
[OAc⁻]o= 0.1 × (50/100) = 0.05 M
The hydrolysis of OAc⁻ :
OAc⁻(aq)+ H2O(l) ⇌ HOAc(aq) + OH⁻(aq) .. Kh
At eqm :
[HOAc] = [OH⁻] =y M
[OAc⁻] =(0.05 - y) M ≈ 0.05 M (Assume that 0.05 >> y)
Kh = [HOAc][OH⁻]/[OAc⁻]
y²/0.05 = (1 × 10⁻¹⁴)/(1.8 × 10⁻⁵)
y = 5.3 × 10⁻⁶
pOH = -log(5.3 × 10⁻⁶) = 5.3
pH = 14 - pOH = 14 - 5.3 = 8.7
2015-01-16 9:41 am
我絕對同意 冷眼旁觀 的觀點！
2015-01-14 7:51 am
「大學」嗎？，連本應獨立自主學習的大學生也要這樣問功課，呵呵…
2015-01-13 8:38 pm
第一題就條件不夠，沒有Ka(HCN)值！當然查表可知
2015-01-13 12:43 pm
可分開嗎?? 用其他方法之類的
我很急需@@"
我很急需@@"
2015-01-13 10:56 am
奇摩答案限 2000 字。而八條題目，若列式詳答，250 字絕不夠用！
2015-01-14 12:09:55 補充：
以前說是大學中學化，現在是大學小學化了。
2015-01-14 12:09:55 補充：
以前說是大學中學化，現在是大學小學化了。
收錄日期: 2021-04-15 17:58:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150112000016KK05831