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

已更新項目:

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?

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

我是這樣算的... 請問哪裡不對?

3 個解答

評分
  • 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.

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