Takuko
Lv 7
Takuko 發問時間： 科學化學 · 1 0 年前

# Acid-base analytical questions

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

To bdref43 大大~

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

FHA > [HA] ?

and also,

2 個已更新項目:

missing a word

i meant,

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

3 個已更新項目:

(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

### 3 個解答

• 1 0 年前
最佳解答

A1：

由弱酸HA的解離方程式

................HA........&larr;&rarr;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！

參考資料： 剛好正在上高二的酸鹼鹽XD, 希望有幫到你的忙唷！(熬夜打...媽媽在催了)
• Takuko
Lv 7
1 0 年前

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&#39;m missing....

• ?
Lv 7
1 0 年前

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.