1) 25.0cm^3 of a sodium carbonate solution was titrated against a 1.0 M hydrochloric acid for complete neutralization. Methyl orange is used as the indicator for the titration. The following shows the results of titration.
Titration 1: initial reading = 0.1cm^3 , final reading = 18.5cm^3
Titration 2: initial reading = 18.5cm^3 , final reading = 36.0cm^3
Titration 3: initial reading = 0.2cm^3 , final reading = 17.6cm^3
Titration 4: initial reading = 17.6cm^3 , final reading = 35.1cm^3
(The reading is the burette reading)
(Data in titration 1 is NOT necessary for question(b) below)
a) State the colour change at the end point of titration.
b) Calculate the average volume of the given hydrochloric acid required for neutralization in cm^3.
c) Write a chemical equation for the reaction.
d) Calculate the molarity of the sodium carbonate solution.
e) Calculate the concentration of the sodium carbonate solution in g/dm^3.
2) 19.76g of an impure sample of anhydrous sodium carbonate is dissolved in water. The solution is made up to 250.0 cm^3 of the solution is titrated by 1.50 M of hydrochloric acid and methyl orange is used as the indicator. The results are shown in the following.
Titration 1: initial reading = 0.2cm^3 , final reading = 22.6cm^3
Titration 2: initial reading = 22.6cm^3 , final reading = 44.2cm^3
Titration 3: initial reading = 0.1cm^3 , final reading = 21.6cm^3
Titration 4: initial reading = 21.6cm^3 , final reading = 43.0cm^3
a) State the colour change at the end point of titration.
b) Calculate the average volume of the given hydrochloric acid required for neutralization in cm^3.
c) Calculate the percentage purity of anhydrous sodium carbonate in the sample.
(Assume that the impurity of sodium carbonate has no reaction with hydrochloric acid.)
Calculation about titration
2008-05-24 2:29 am
回答 (3)
2008-05-24 10:29 am
✔ 最佳答案
1.(a) Colour change at the end point is from yellow to orange.
(b)
Volume of HCl used in titration 2 = 26.0 - 18.5 = 17.5 cm3
Volume of HCl used in titration 3 = 17.6 - 0.2 = 17.4 cm3
Volume of HCl used in titration 4 = 35.1 - 17.6 = 17.5 cm3
Average of HCl used = (17.5 + 17.4 + 17.5) / 3 = 17.47 cm3
(c)
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
(d)
Mole ratio Na2CO3 : HCl = 1 : 2
No. of moles of HCl = MV = 1 x (17.47/1000) = 0.01747 mol
No. of moles of Na2CO3 = 0.01747 x (1/2) = 0.008735 mol
Molarity of Na2CO3 = mol/V = 0.008735/(25/1000) = 0.3494 mol dm-3
(e)
Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g mol-1
Concentration of Na2CO3 = 0.3494 mol dm-3
Concentration of Na2CO3 = (0.3494 mol dm-3) x (106 g mol-1)
Concentration of Na2CO3 = 37.04 g dm-3
=====
(2)
In the question,
「...... made up to 250.0 cm3 of the solution is titrated by」
should be 「......made up to 250.0 cm3. 25 cm3 of the solution is titrated by ......」
(a) Colour change at the end point is from yellow to orange.
(b)
Volume of HCl used in titration 1 = 22.6 - 0.2 = 22.4 cm3 (rejected)
Volume of HCl used in titration 2 = 44.2 - 22.6 = 21.6 cm3
Volume of HCl used in titration 3 = 21.6 - 0.1 = 21.5 cm3
Volume of HCl used in titration 4 = 43.0 - 21.6 = 21.4 cm3
Average of HCl used = (21.6 + 21.5 + 21.4) / 3 = 21.5 cm3
(c)
Consider the titration:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Mole ratio Na2CO3 : HCl = 1 : 2
No. of moles of HCl used = MV = 1.5 x (21.5/1000) = 0.03225 mol
No. of moles of Na2CO3 used = 0.03225 x (1/2) = 0.01613 mol
Molarity of Na2CO3 = mol/V = 0.01613/(25/1000) = 0.6452 mol dm-3
Consider the 250 cm3:
No. of moles of Na2CO3 = MV = 0.6452 x (250/1000) = 0.1613 mol
Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g mol-1
Mass of Na2CO3 = mol x (molar mass) = 0.1613 x 106 = 17.10 g
% by mass of Na2CO3 = (17.10/19.76) x 100% = 86.54(%)
2008-05-24 02:30:26 補充：
In (2), if all 250 cm^3 of the solution is used in titration, it becomes:
No. of moles of Na2CO3 in the solution = 0.03225 x (1/2) = 0.01613 mol
Mass of Na2CO3 = mol x (molar mass) = 0.01613 x 106 = 1.710 g
% by mass of Na2CO3 = (1.710/19.76) x 100% = 8.654(%)
2008-05-24 5:14 pm
To: ncy_0916
You have some careless mistake.
「d)the no. mol of HCl = MV
= (1.0)(17.47/1000)
= 0.01747mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = 0.0349mol 」
No. of mol should be 0.01747/2 = 0.008735.
You have some careless mistake.
「d)the no. mol of HCl = MV
= (1.0)(17.47/1000)
= 0.01747mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = 0.0349mol 」
No. of mol should be 0.01747/2 = 0.008735.
2008-05-24 3:50 am
1a)the colour change at the end point is from yellow to orange.
b)the volume of titration 2 = 36.0﹣18.5 = 17.5cm3
the volume of titration 3 = 17.6﹣0.2 = 17.4cm3
the volume of titration 4 = 35.1﹣17.6 = 17.5cm3
the average volume of the given hydrochloric acid required for neutralization
= (17.5 + 17.4 + 17.5) / 3
= 17.47cm3
c)Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
d)the no. mol of HCl = MV
= (1.0)(17.47/1000)
= 0.01747mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = 0.0349mol
the molarity of the sodium carbonate solution = no. mol / volume
= 0.0349 / (25.0/1000)
= 1.40M
e)no. mol = mass / molar mass
∴the mass of Na2CO3 = (0.0349)(2X23 + 12 + 3X16)
= 3.70g
∴the concentration of the sodium carbonate solution
= 3.70 / (25.0/1000)
= 148gdm-3
2a)the colour change at the end point is from yellow to orange
b)the volume of titration 2 = 44.2﹣22.6 = 21.6cm3
the volume of titration 3 = 21.6﹣0.1 = 21.5cm3
the volume of titration 4 = 43.0﹣21.6 = 21.4cm3
the average volume of the given hydrochloric acid required for neutralization
= (21.6 + 21.5 + 21.4) / 3
= 21.5cm3
c)Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
the molar mass of the sample of anhydrous sodium carbonate
= 2X23 + 12 + 3X16 + 10(2X1 + 16)
= 286 gmol-1
the no. mol of HCl = (21.5)(1.50/1000) = 0.03225mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = (1/2)(0.03225) = 0.016125mol
the mass of sodium carbonate in the sample
= (0.016125)(286)
= 4.61175g
∴the percentage purity of anhydrous sodium carbonate in the sample
= (4.61175 / 19.76)(100%)
= 23.3%
b)the volume of titration 2 = 36.0﹣18.5 = 17.5cm3
the volume of titration 3 = 17.6﹣0.2 = 17.4cm3
the volume of titration 4 = 35.1﹣17.6 = 17.5cm3
the average volume of the given hydrochloric acid required for neutralization
= (17.5 + 17.4 + 17.5) / 3
= 17.47cm3
c)Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
d)the no. mol of HCl = MV
= (1.0)(17.47/1000)
= 0.01747mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = 0.0349mol
the molarity of the sodium carbonate solution = no. mol / volume
= 0.0349 / (25.0/1000)
= 1.40M
e)no. mol = mass / molar mass
∴the mass of Na2CO3 = (0.0349)(2X23 + 12 + 3X16)
= 3.70g
∴the concentration of the sodium carbonate solution
= 3.70 / (25.0/1000)
= 148gdm-3
2a)the colour change at the end point is from yellow to orange
b)the volume of titration 2 = 44.2﹣22.6 = 21.6cm3
the volume of titration 3 = 21.6﹣0.1 = 21.5cm3
the volume of titration 4 = 43.0﹣21.6 = 21.4cm3
the average volume of the given hydrochloric acid required for neutralization
= (21.6 + 21.5 + 21.4) / 3
= 21.5cm3
c)Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
the molar mass of the sample of anhydrous sodium carbonate
= 2X23 + 12 + 3X16 + 10(2X1 + 16)
= 286 gmol-1
the no. mol of HCl = (21.5)(1.50/1000) = 0.03225mol
By the chemical equation,we need 1mol of Na2CO3 to neutralize 2mol og HCl.
∴the no. mol of Na2CO3 = (1/2)(0.03225) = 0.016125mol
the mass of sodium carbonate in the sample
= (0.016125)(286)
= 4.61175g
∴the percentage purity of anhydrous sodium carbonate in the sample
= (4.61175 / 19.76)(100%)
= 23.3%
收錄日期: 2021-04-23 23:07:24
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