Acid-base analytical questions

Question

I've got three relatively easy problems that I don't know how to solve...
Can someone explain it to me? thx
Q1...The pH of a weak acid solution is predicted by the approximate formula [H ] = squre root of (Ka * FHA). The predicted value of [H ] is found to be about 0.5% of FHA. Compared to the pH calculated, the true pH is (to two decimal places):
a) significantly higher
b) negligibly higher
c) significantly lower
d) negligibly lower <--- i think it's this one but i don't know how to explain/calculate it
Q2... The pH of 1.0 x 10-6 M NaOH (aq) lies between:
a) 7.96 and 8.00
b) 7.96 and 8.04
c) 8.04 and 8.10
d) 8.00 and 8.04
Q3... A weak acid is titrated with strong base. The pH is recorded one-third of the way to the equivalence point, then again two-thirds of the way to the equivalence point. The difference between the two pH values will be:
a) 0.30
b) 0.50
c) 0.60
d) 0.66

謝回答~

To bdref43 大大~

why is FHA=[HA]-[H ] ??

FHA > [HA] ?

and also,
about Q2
照芒果大的算法, ph = 8.004....

為什麼 d) 8.00 ~ 8.04 不行?

missing a word

i meant,

shouldn't FHA > [HA]? and not the other way around...?

(10^-7 - x)(1.1x10^-6 -x) = 10^-14

x^2 - 1.2x10^-6 10^-13 = 0

x = 9.0098 x 10^-8 跟 1.1099 x 10^-6

[H ] = (10^-7 - x) 所以用 9.0098 x 10^-8

[H ] = 9.90195 x 10^-9

pH = 8.004279

我是這樣算的... 請問哪裡不對?

Answer 1

A1：
由弱酸HA的解離方程式
................HA........←→H﹢＋A¯
反應前...FHA
反應中...-x................... x..... x
______________________________
反應後(FHA-x)..............x.......x

得到 (x˙x)／(FHA-x)=Ka

此時若假設 x遠小於FHA而省略(FHA-x)中的x

則得到題幹敘述中的公式，

若不省略，所得到的解將會比省略得到的解小
(這要用一些數學伎倆觀察)
換句話說，真正的[H ]會比公式算出來的小，故pH值較大

因為HA是弱酸，(x˙x)／(FHA-x)≒(x˙x)／FHA，

pH只有大一點點而已，

故答案是b) negligibly higher

2007-03-08 01:05:34 補充：
A2:
此題要考慮水的自身解離
................H2O ←→ H﹢＋ OH¯
反應前........................10^-7.......1.1*10^-6
反應中...........................-x...............-x
__________________________________
反應後......................(10^-7-x) (1.1*10^-6- x)

2007-03-08 01:05:40 補充：
此時(10^-7-x) ˙(1.1*10^-6- x)＝Kw=10^-14

加油吧！這時就要仰賴您按計算機的能力了！

2007-03-08 01:07:07 補充：
A3：
不妨假設有3mol.的弱酸HA，(體積為V)

如果把開始滴定至到達當量點當做全程，
(照滴入的強簡體積分段)

滴定到達三分之一時(設為狀況A)，還有2mol.的HA和1mol.的A¯


滴定到達三分之二時(設為狀況B)，還有1mol.的HA和2mol.的A¯

2007-03-08 01:07:31 補充：
According to 海德莫方程式

狀況A的pH＝pKa log{[A¯]／[HA]}--------(1)

狀況B的pH'＝pKa log{[A¯]'／[HA]'}-------(2)

由(2)－(1)：

pH'－pH＝log2－log0.5＝log2＋log2≒0.3010＋0.3010＝0.6020#

答案選(C)

2007-03-08 22:51:23 補充：
我算出來答案卻是8.0453左右耶，所以答案應是c) 8.04 and 8.10

答案是較大的那個解唷！(pH值大於七那一個)

2007-03-09 22:31:27 補充：
不好意思，是我自己按錯了，

在此提供一個比較投機的算法：
................H2O ←→ H﹢＋ OH¯
反應前........................0...............10^-6
反應中........................... a............... a
__________________________________
反應後.............................a..........10^-6 a

2007-03-09 22:31:40 補充：
所以a^2＋10^-6*a－10^-14＝0

a＝9.902*10^-9
pH＝8.00427
您算的沒錯！

2007-03-09 22:43:49 補充：
如果Q2是單選題的話
假若(A)正確則(B)必正確
假若(B)正確則A)、(D)其一必正確
假若(D)正確則B)必正確
選(A)、(B)、(D)皆矛盾，依邏輯就能選出(C)
可是算出來竟然不是在選項(C)的範圍之內。
(答案我已經再算過很多次，應是8.00427沒錯)
抱歉，這點我可能也無法解釋！

Answer 2

thx~

but i still have the dillema in Q2 in choosing b or d..

although the more i think about it, c does make sense for sake of it being exam questions

i wonder what i'm missing....

Answer 3

You are right. I misplaced the sign. FHA=[HA]plus [H+]. This will change the result to [HA]=0.995FHA and the true [H+] will be (0.995KaFHA)^0.5 and pH is 0.0011-0.5log(KaFHA) is negligibly higher.

2007-03-09 02:54:06 補充：
I gave the answer for Q2 too quick. If you consider charge balance [Na]plus[H]=[OH], [Na]=10^-6, [OH]=10^-14/[H]. Therefore [H]^2plus10^-6[H]-10^-14=0, [H]=9.9*10^-9, pH=8.004

Since I made a mistake in calculations, I should pull out my answers. The worst thing to do is to mislead you.