Question one
It is found experimentally that the amount of CaS04 that will dissolve in 500 ml of water
at 25 °C is 0.204 g. Calculate the value of Ks (CaS04) at 25°C. M (CaS04) = 136.1gmol”1
Show all of your working. Quote your final answer to three significant figures.
Solution
The mass of CaSO4 soluble in 1000ml of water at 25C is computed as follows;
500ml = 0.204g
1000 ml= 0.204 × 2
=0.408g at 25 °C.
The number of moles present in 1000ml is computed as follows;
Number of moles = Mass/Formula mass
= 0.408g/118
= 0.003458 mole
Utilizing the mole and the concentration in g/mol, we get the value of M.
If 1mole = 136.1g
0.003458 mole = 1× 0.003458/136.1
= 4.71x 10-1
M is therefore equivalent to 4.71x 10-1
Question two
The value of Ks (Ag2 Cr04) is 3.00 x 10″12 at 25°C. Calculate the maximum mass of solid AgCr04 that can be dissolved in 100 ml of distilled water at 25°C. Show all of your working. Quote your final answer to three significant figures. /W (Ag2 Cr0 4) = 331.8gmol’1.
Solution
The mass of AgCr04dissolved in 1000ml at 25C is 3.318g
Ksp = [Ag+]2[Cr04-]
= 3.00 x 10-12
Ag2 Cr04 (s) D2 Ag+ (aq) + CrO42- (aq)
3.00 x 10-12 2x x
X3 = 3.00 x 10-12
X= 1.442 x 10-4 moles
No of mole present 1.442 x 10-4 moles
Mass of the AgCr04 = Number of moles× Formula mass
= 1.442 x 10-4× 331.8gmol’1
= 0.04785g
Therefore, the maximum mass of AgCr04 within 100 ml of condensed water at 25°C is 0.04785g
Question three
A saturated aqueous solution of calcium hydroxide, limewater, contains 0.740 g of solid
Ca (OH) 2 per litre at 25°C. Calculate the value of Ks (Ca (OH)2) at 25°C. M (Ca (OH) 2) = 74.1gmor1. Show all of your working. Quote your final answer to three significant figures.
Solution
From the equation of the reaction, the number of moles that take part in the reaction determines formation of precipitates. In this case, a precipitate does not form because the two reagents are in equilibrium.
The mass of Ca (OH)2 dissolved in 1000ml at 25C is 0.740g.
No of mole present 0.016mol
Therefore M = 74.1g/mol 0.016
0.00021
M= 2.17 x 10-4
Question Four
Will a precipitate of PbS04 occur when 0.0390 g of Pb(N03)2is dissolved in 500 ml of
1.01 x 10″4 mol L”1 sulfuric acid (H2SO4) solution? Show all of your working to justify your answer. Ks (PbS04) = 1.30 x 10″8. M (Pb (N03)2) = 331.2gmol”1.
Solution
A precipitate does not form. The Pb (NO3)2 is in excess.
Mole of PbSO4= 1.30X 10-8 * 232= 0.301x 10-5
Mole of PbNO available= 0.039/331.2 = 1.17x 10-4 mol
Mole of H2SO4 available = 1.01x 10-8*0.5 = 5.5 x 10-9 mol
Question five
Will a precipitate of Ag2C03 occur when 50.0 ml of 0.00342 mol L”1 Na2C03 solution is
Added to 50.0 ml of 1.15 x 10″4 mol L”1 AgN03 solution? Show all of your working to justify your answer. Ks (Ag2C03 ) = 8.00 x 10-12
Solution
Mole of Sodium Carbonate available 1.71 x 10-5 mol
Moles of silver Nitrate available 5.75x 10-6 mol
The two combine in a ratio of 2:1
Therefore the number of moles of AgNO3 should be 2.375 x 10-6mol.
The number of mole of Ag2CO3 that completely dissolves is 8.0 x 10-12
Therefore a precipitate forms.
Question Six
20.0ml of 0.0315molL-1 Mg (NO3) are added to 60.0ml of 0.00162mol/L NaOH. Will a precipitate of Mg (OH) 2 form.
Solution
No of moles of Magnesium Nitrate available = 6.3 x 10-4mol
Number of moles of Sodium Hydroxide available= 9.72 x10-5 mol
The mole ratio is 1: 2
Therefore, the number of moles needed for a precipitates to occur are not available.
