SS1 CHEMISTRY THIRD TERM EXAMINATION

Example 1:
A is a solution of tetraoxosulphate {vi} acid.
B is a solution containing 0.0500 mole of anhydrous Na2CO3 per dm3.
(a) Put A in the burette and titrate 20.00 or 25.00 cm3 portions of B using methyl orange as the indicator. Record the size of your pipette. Tabulate the burette readings, and calculate the average volume of the acid used.
(b) From your result and data provided, calculate the
(i) amount of Na2CO3 IN 25.00 CM3 OF B used
(ii) concentration of A in moldm-3
(iii) concentration of A in gdm-3
(iv) number of hydrogen ions in 1.00dm3 of A. {Avogadro number = 6.02 × 1023 mol-1}
The equation of reaction is: H2SO4{aq} + Na2CO3{aq}   →   Na2SO4 {aq} + H2O + CO2
{H = 1, O = 16, S = 32}
Solution:
A volume of pipette: 25cm-3
Titration Results {Hypothetical Data}

Burette reading Icm3 IIcm3 IIIcm3
Final 24.75 49.15 25.7
Initial 0 24.75 1.35
Volume of acid used 24.75 24.4 24.35

Average volume of acid used from titrations II and III:
24.40+24.352=48.752=24.38cm3
NOTE: Only the titre values from titrations I and II can be used in averaging, since they are within 0.20cm3 of each other. …Rough of first titre can also be used in averaging,  if it is within 0.20cm3 of any other titre value, and is not crossed.—–Do not round up 24.38cm3 to 24.40cm3
(b) To calculate the amount of Na2CO3 in 25.00cm3 Given: conc of  B = 0.050moldm3 : Volume = 25/1000 dm3

Amount =Concmoldm3×Volumedm3=0.050×251000=0.00125mol

To calculate the concentration of A in moldm3: The various titration variables are:
CA = xmoldm3; VA = 24.38cm3; Na = 1, CB = 0.050moldm3 VB = 25cm3: nB = 1
Method 1: Proportion method {from the first principle}
From the balanced equation of reaction:
1mol of Na2CO3 = 1 mol of H2SO4
∴ 0.00125mol of Na2CO = 0.00125mol of H2SO4
i.e. 24.38cm3 of A contained 0.00125mol of H2SO4
∴ 1000cm3 of A contained 0.00125×100024.38mol=0.0513mol. Hence, concentration of A is 0.0513mol moldm3.
Method 2:
Mathematics Formula Method
From the data above, it is safe to use the mole ratio expression, in order to calculate the concentration CA of A, which the required variable.
CAVACBVB=nAnB
Substituting;
CA×24.380.050×25=11
Making CA the subject of the formula
∴ CA=1×0.050×2524.38×1=0.0513moldm−3
(iii) To calculate the concentration of A in gdm-3:
Using conc. {gdm-3} × Molar mass {gmol-1}
Concentration of H2SO4, moldm-3 = 0.0513moldm-
Molar mass H2SO4, = 2 {1.0} + 32.0 + 4{16.0
= 2.0 + 32.0 + 64.0 = 98.0gmol-1
Substituting; Mass conc = 0.0513 × 98 = 5.0274gdm-3.
= 5.03gdm-3 {3 sig fig.}
(iv) Number of hydrogen ions in 1.00dm3 of A
1 dm3 of A contained 0.0513mol of H2SO4.
H2SO4 ionizes in water completely thus:
H2SO4 (a q)   →   2H+SO42-(a q)
1mol  2mol
From the equation;
1 mole of H2SO4 produces 2 × 0.0513 moles of H+ = 0.103 mol 0f H+
But 1 mole of H+ contains 6.02 × 1023 ions;
Therefore, 0.103 × 6.02 × 1023 ions = 6.02 × 1023 ions. {3 sig fig.}
1. Determination of Relative Molar Mass
In the determination of relative molar mass of an acid or bases by titration, the concentration of both acid and base, at least in gram per dm3, will be provided together with the balanced equation of reaction so as to establish the mole ratio
Example 1:
A contains 1.6g of HNO3 in 250cm3 of solution
B contains 10gdm-3 of XHCO3
25cm3 portion of B requires an average of 24.90cm3 of A for complete neutralization. Calculate the
⦁ Concentration of acid in moldm-3
⦁ Concentration of XHCO3 in B in moldm-3
⦁ Mass of XHCO3
⦁ Value of X
Equation of the Reaction
HNO3(aq) + XHCO3(aq) → XNO3(aq) + CO2(g) + H2O(l)
(H = 1; C = 12; O = 16)
Solution
Ca = ? Mole per dm3; Va = 24.90cm3; na =1
Cb = ? Mole per dm3; Vb = 25cm3; nb = 1
(1) To calculate the molar concentration of A.
The given mass concentration of A is
1.6g of HNO3 in 250cm3
i.e. 250cm3 of the solution contained 1.6g of acid
Therefore, 1000cm3 of A contained
1.0×1000250=6.40gdm−3
i.e. the mass concentration of A = 6.40gdm-3
But molar concentration = mass concentrationmolar mass
Molar mass of HNO3 = (1 × 1) + (14 × 1) + (16 × 3) = 1 + 14 + 48 = 63gmol-1
Molar conc Ca=6.463=0.102moldm−3
(2) To calculate the molar concentration of B
Mass conc. Of XHCO3 in B = 10.0gdm-3 (given)
Since the value of X in the base XHCO3 is not known, the mole ratio expression must be used in order to find its molar concentration,
CaVana=CbVbnb
Substituting,
0.102×24.91=Cb×251
Making Cb the subject of the formula
Cb=0.102×24.925=0.102moldm−3
(3) To find the mass of X in XHCO3
molar mass
=mass concmolar conc=10.0gdm−30.102moldm−3=98.0gmol−1
(4) To find the value of X in XHCO3
Molar mass of XHCO3 = 98.0gmol-1
i.e  X + 1 + 12 + (16 × 3) = 98
X + 1 + 12 + 48 = 98
X + 61 = 98
X = 98 − 61
Relative atomic mass of X is  X = 37 (no unit)
2. Determination of Mole Ratio
In order to determine the mole ratio of acid to base (or base to acid) by titration, the solution of the acid and the base provided for the titration must be known concentrations (standard solution). However, the equation of reaction will not be provided.
Solved Problem 1:
A is a solution of 0.0500moldm-3 hydrochloric acid
B is 0.0250moldm-3 of a trioxocarbonate (iv) solution
25.00cm3 portions of B required an average of 24.60cm3 of A for complete neutralization, using methyl orange as the indicator
(a) Calculate the:
(i) amount of acid in the average volume of A used
(ii) amount of trioxocarbonate (IV) in the volume of B used
(iii) mole ratio of the acid to trioxocarbonate (IV) solution in the reaction, express your answer as a whole number ratio of one
(b) State whether the PH of the following would be equal to 7, greater than 7 or less than 7
(i) solution A
(ii) solution B
(iii) titration mixture of A and B before end point
Solution:
(a) (i) To calculate the amount of acid in A used
Amount of the acid = molar conc of A × volume in dm3.
=0.050×24.601000mol=0.00123mol
(a) (ii) To calculate the amount of trioxocarbonate (iV) in B used
Amount of Base = molar conc. of B × volume in dm3
=0.025×251000mol=0.000625mol
(a) (iii) Mole ratio of acid to trioxocarbonate (iV)
Mole ratio of acid to base, A : B = 0.00123 : 0.000625
A:B=0.001230.000625:0.0006250.000625A:B=2:1
(i) PH of A is less than 7
(ii) PH of B is greater than 7
(iii) PH is titration mixture before end point is greater than 7
Solved Problem 2:
In an acid – base titration, 24.80cm3 of 0.05000moldm-3 of a mineral acid Y neutralized 25.00cm3 of a solution containing 5.83g of Na2 CO3 per dm3.
(a) From the information given above calculate the:
(i) amount (in moles) of acid Y used
(ii) amount in moles of Na2CO3 used
(iii) mole ratio of the acid to Na2CO3 in the titration
(b) (i) What is the basicity of Y?
(ii) Suggest what Y could be. Give reason for your answer.
(iii) From your answer, write a balance equation of the reaction
Solution:
(a) (i) Amount ny = conc (moldm-3) × volume (dm3)
Amount in mole of Na2CO3 used
First, calculate the concentration of Na2CO3 in moldm-3
Molar mass of Na2CO3 =2(23.0)+12+3(16)=46+12+48=106gmol−1
Conc of Na2CO3 =mass concentrationmolar mass=5.83106=0.0550moldm−3
Amount of Na2CO3,
nz=0.0550×251000=0.00138mol
(iii) Mole ratio of the acid to Na2CO3 in the titration
Mole ratio of the acid to Na2CO3 = ny : nz = 0.00124 : 0.00138
ny:nz=0.001240.00124:0.001380.00124ny:nz=1:1
(b) (i) The basicity of Y is Y since one mole of Y requires one mole of Na2CO3 .
(ii) Y is H2SO4. Reason: one mole of the acid Y produces two mole of H+.
H2SO4 + NaCO3 → Na2SO4 + H2O + CO2
(C = 12.0, O = 16.0, Na = 23.0)