酸鹼滴定的計算
摩爾濃度與容量分析
溶液的濃度
以下有兩個方法表達溶液的濃度
○,1濃度以 g dm-3 為單位
溶液的濃度(g dm-3) =
(1dm3=1000cm3=1L)
Example:
當 5 克的 NaCl 浴於 200mL 的蒸餾水,
溶液的濃度 (g dm-3) =
○,2濃度以 mol dm-3 為單位
溶液的濃度 (mol dm-3) =
Example:
當 5 克的 NaCl 浴於 200mL 的蒸餾水,
溶液的濃度 (mol dm-3) =
Remarks:
1. 當濃度以 mol dm-3 為單位時，我們會稱之為摩爾濃度
2. mol dm-3 可以用 M 當作單位
關於溶液濃度的簡單計算
練習一:
57.2 克的水合碳酸鈉 Na2CO3．10H2O 被溶於 50cm3 的蒸餾水。請計算其
摩爾濃度?
(R.A.M.: Na=23.0, C=12.0, O=16.0) Ans=4.0M
練習二:
若要準備 250cm3 0.5M 的硫酸銅溶液, 我們需要多少克的硫酸銅?
(R.A.M.: Cu=63.5, S=32.0, O=16.0) Ans= 19.94g
練習三:
濃硝酸的樣本密度是 1.37g cm-3，而硝酸於樣本中的質量百分比是 65.4%。
請計算硝酸於樣本的摩爾濃度。
(R.A.M.: H=1.0, N=14.0, O=16.0) Ans=14.2M
Class Exercise:
濃硫酸樣本的密度是 1.83g cm-3 而硫酸於樣本中的質量百分比是 98.0%
硫酸樣本的濃度是多少？
(R.A.M: H=1.0, O=16.0, S=32.1)
A. 18.0M
B. 18.3M
C. 18.7M
D. 19.8M
稀釋
稀釋過程中所加入的水並不會影響溶質的摩爾數，因為水並不含有任何溶質。
溶質在稀釋前的摩爾數=溶質在稀釋後的摩爾數
n1=n2
摩爾濃度 x 稀釋前的容量=摩爾濃度 x 稀釋後的容量
C1 V1 = C2 V2
稀釋後的容量= 溶液的最終濃度
但並不等於所加入的水的容量
Calculations involving dilution
Example 1:
You want to dilute a solution to 1.5M sulphuric acid to obtain 250cm3
of 0.10 M sulphuric acid. Calculate the volume of 1.5M sulphuric acid
needed. (Ans=16.67cm3)
Example 2:
50cm3 of distilled water is added to 750cm3 of 1.5M sodium chloride
solution. What is the molarity of the diluted solution?
Example 3:
A student adds water to 30.0cm3 of 8.25M Na2CO3 to make it exactly
2.5M. What is the volume of water added?
(Ans=69cm3)
Class Exercise :
The percentage by mass of M2CO3 in a sample is 65%. 20.0g of the sample was added
to the 35cm3 of water. It was found that the concentration of metal cation is 5.38M.
What is element M?
Class Exercise1:
When 2M of HNO3 was allowed to react separately with 2M of NaOH and 2M of
KOH, the experiment results and the temperature change were recorded as
follows:
Expt. No
1
2
Volume of acid
Volume of alkali
50 cm3 of 2M HNO3 50 cm3 of 2M NaOH
200 cm3 of 2M
HNO3
200 cm3 of 2M
NaOH
Rise in temperature
T1℃
T2℃
Which of the followings is correct?
A. T1=T2
B. 2TI=T2
C. 4T1=T2
D. 8T1=T2
Class Exercise 2:
Which of the following statements concerning 200.0cm3 of 2M nitric acid and
200.0cm3 2M ethanoic acid is/are correct?
(1) They have similar electrical conductivity.
(2) They react with sodium hydrogencarbonate at the same rate.
(3) They require the same volume of 2M sodium hydroxide solution for complete
neutralisation.
A.
B.
C.
D.
(1) only
(3) only
(1) and (3) only
(1) and (3) only
Acid-alkali titration: a method to determine the concentration of a
solution of an acid(alkali) by using a solution of an alkali(acid) of known
concentration
Steps:
(Assume the solution of unknown concentration is solution A and that
of know concentration is solution B)
1. Wash a 25cm3 pipette with distilled water and then with solution A.
Transfer 25cm3 of solution A to a conical flask using the pipette.
2. Add a few drops of an alkali to solution A.
3. Wash the burette with distilled water and then with solution B. Fill
the burette with solution B. Fill the burette with solution B. Ensure
no bubbles trapped in the burette. Record the initial burette reading.
4. Set up the apparatus as follows:
5. Run solution B from the burette into solution A until the indicator
in the conical flask just changes colour (this point is called end
point).
6. Record the final burette reading and calculate the volume of
solution B used ( Final reading - Initial reading)
7. Repeat 2 to 3 times
From the results, we can calculate the concentration of solution A.
Important points to remember
Apparatus
Substances used to wash
the apparatus before the
experiment
Reason
Burette
(i) Distilled water
(ii) Titrant
(i) Remove water-soluble
impurities
(ii) Prevent the dilution of
titrant by the distilled water
prevent in the burette.
Pipette
(i) Distilled water
(i) Remove water-soluble
(ii) Sample solution
impurities
(ii) Prevent the dilution of
sample by the distilled water
present in the pipette.
Otherwise, the number of
moles of sample transferred
to the conical flask will be
reduced.
Distilled water ONLY
If the conical flask is washed
Conical flask
with the sample solution, the
few drops of sample solution
remain in the conical flask
will react with the titrant
during the titration. This will
cause an over-estimation of
titrant required in the
titration.
Choice of indicator
Two indicators, methyl orange and phenolphthalein are commonly used
in acid-alkali titrations.
Indicator
Colour
In acidic solution
In alkaline solution
Methyl orange
Red
Yellow
Phenolphthalein
Colourless
Red/Pink
1. Methyl orange can be used when the titration involves a strong acid.
2. Phenolphthalein can be used when the titration involves a strong
alkali.
Acid-Alkali titration
Example
Suitable indicator
Strong acid and Strong
alkali
HCl and NaOH
Methyl orange /
Phenolphthalein
Strong acid and weak
alkali
H2SO4 and NH3
Methyl orange
Weak acid and strong
alkali
CH3COOH and KOH
Phenolphthalein
Weak acid and weak
alkali
CH3COOH and NH3
None
Calculations involving titrations
Example 1:
The concentration of a NaOH solution is 2.76M. 25cm3 of this solution
requires 33.0cm3 of nitric acid for complete reaction. What is the
molarity of the acid?
(Ans=2.1M)
Example 2:
22.68g of pure ethanedioic acid crystals (COOH)2．2H2O is dissolved in
250 cm3 distilled water. 25.0 cm3 of this solution is titrated against
0.400M of NaOH solution. What is the volume of NaOH solution
required for complete neutralization?
(R.A.M: H=1.0, C=12.0, O=16.0)
Example 3:
The following table shows the experimental results for the titration of
25.0cm3 ammonia against 0.20M sulphuric acid.
Burette
Reading(cm3)
Titration
1st
2nd
3rd
4th
Final
Reading
22.8
42.7
22.1
41.8
Initial
Reading
0
22.8
2.0
21.8
a) Suggest a suitable indicator, what is the colour change at the end
point?
b) Calculate a reasonable average for the volume of sulphuric acid
required to completely neutralize 25.0 cm3 of the alkali.
c) Calculate the molarity of the ammonia solution
Example 4:
4.15g of hydrated sodium carbonate Na2CO3．aH2O is dissolved in
250cm3 distilled water. 25cm3 of this solution requires 29.0 cm3 of
0.0500M sulphuric acid for complete reaction. Find the value of a.
(R.A.M.: H=1.0, C=12.0, O=16.0, Na=23.0)
Example 5:
The molar mass of ascorbic acid is 176.0gmol-1. 2.20g of ascorbic acid is
dissolved in 25cm3 distilled water. This solution requires 5.0 cm3 of
2.5M KOH solution for complete neutralization. Calculate the basicity
of the acid.
Example 6:
5.30g of an impure sample of ethanedioic acid (COOH)2．2H2O is
dissolved in 25.0 cm3 distilled water. This solution requires 40.0cm3 of
2.00 KOH solution for complete neutralization. Calculate the
percentage purity of ethanedioic acid in the sample
Example 7:
The following experiment is performed to determine the percentage by
mass of calcium carbonate in a sample:
Step 1
11.0g of sample was placed in a beaker
Step 2
120cm3 of 2.00M nitric acid was added to react with the
sample
Step 3
The excess acid was titrated against 1.60M sodium
hydroxide solution with phenolphthalein as indicator.
17.5cm3 of the alkali were required to reach the end point
Calculate the percentage by mass of calcium carbonate in the sample.
(R.A.M.: Ca=40.1, C=12.0, O=16.0, H=1.0)
Example 8:
In order to find out the value of x and y in the sample (formula = MgCl2
xKCl y K2SO4), a student conducted the following experiments:
(1) 10g sample was dissolved in 50.0cm3 of H2O. When excess sodium
hydroxide solution was added to this solution, 1.392g of white
precipitate was obtained.
(2) 5g sample was dissolved in 50cm3 H2O. When excess calcium
bromide solution was added to this solution, 1.626g white percipitate
was obtained.
Calculate the value of x and y.
(R.A.M:H=1.0, O=16.0, Mg=24.3, S=32.1, Cl=35.5, K=39.1, Ca=40.1)
Example 9:
A sample contains 21.0% of metal X by mass, which is a group III
element. 19g of the sample was added to 100cm3 of 2M sulphuric acid.
The remaining acid required 45.56cm3 of 5M potassium hydroxide for
complete neutralization. What is metal X?
練習一:
氯化鈉溶液的濃度為 2.76M. 25cm3 的溶液需要 33.0cm3 的硝酸作完全反應。
酸的摩爾濃度是多少?
(Ans: 2.1M)
練習二:
把 22.68g 的純乙二酸晶體 (COOH)2．2H2O 溶於 250cm3 的蒸餾水中。再把
25cm3 的溶液來滴定置於錐形瓶的 0.400M 的氯化鈉溶液。要用多少的氯化鈉溶
液才能完成完全中和作用?
(R.A.M. : H=1.0, C=12.0, O=16.0
(Ans=90cm3)
練習三:
以下表格展示了 25cm3 氨溶液滴定 0.2M 硫酸的實驗結果.
滴定管的讀數
滴定
1st
2nd
3rd
4th
最終讀數
22.8
42.7
22.1
41.8
起始讀數
0
22.8
2.0
21.8
a) 建議一個合適的指示劑，並說明在終點時其顏色的轉變
b) 請計算硫酸與鹼作完全中和作用時所使用的合理平均容量。
c) 請計算氨溶液的摩爾濃度.
練習四:
4.15g 的水合碳酸鈉 Na2CO3．aH2O 溶於 250cm3 的蒸餾水中。而 25cm3 的溶
液需要 29.0cm3 的 0.0500M 硫酸作完全反應。找出 a 的數值。
練習五:
抗壞血酸的摩爾質量為 176.0gmol-1. 2.20g 的抗壞血酸溶於 25cm3 的蒸餾水中，
其溶液與 2.5M 的 KOH 溶液作完作中和作用。請計算酸的鹽基度。
練習六:
5.30g 的含雜質的乙二酸樣本(COOH)2．2H2O 溶於 25.0cm3 的蒸餾水。而溶液
與 40.0cm3 的 2.00M KOH 作中和作用。請計算乙二酸樣本的純度百分比。
(R.A.M: H=1.0, C=12.0, O=16.0)
練習七:
以下實驗目的是找出碳酸鈣樣本的質量百分比
第一步 11.0g 樣本放置在燒杯中
第二步 120cm3 的 2.0M 硝酸加在燒杯中與樣本反應
第三步 把過量的酸滴定置於錐形瓶的 1.60M 氫氧化鈉溶液並而酚酞作指示
劑.使用了 17.5cm3 的鹼令反應達至終點。
請計算碳酸鈣樣本的質量百分比.
(R.A.M: Ca=40.0, C=12.0, O=16.0, H=1.0)
練習八:
一個樣本含有 21%的金屬 X，而這個金屬是屬於第三族。把 19g 的樣本加在
100cm3 的 2M 硫酸中。而剩下的酸再與 45.56cm3 的 5M 氫氧化鉀作完全中和
作用。
金屬 X 是什麼?
練習九:
2g 的氧化銅(II)加在 25cm3 含 6M 酸的樣本中，把溶液稀釋為 250cm3. 25cm3
的溶液與 25cm3 的 1M 氫氧化鈉作完全反應。請找出酸的鹽基度。