Solutions for Exercise (Module: Acid and Bases)
IJSO Training: Acid and Bases
Suggested / Acceptable Solutions
1.Molar mass of NaOH=23 + 16 + 1
=40
Mole of NaOH=10.0/40
=0.25
Molarity of solution=0.25/0.2
=1.25M
2.Moles of organic acid in 25mL solution=2.50/126 x 25/250
=0.00198
Moles of NaOH used=0.1000 x 39.7/1000
=0.00397
Moles of NaOH/Moles of organic acid=0.00397/0.00198
=~2.01
2
The basicity of the organic acid is 2.
3.(i)pH =-log[H+]
=-log[0.01]
=-(-2)
=2
(ii)pH =7since the acid is neutralized by the base.
(iii)After the neutralization reaction,
Concentration of NaOH in the solution =0.01 x 0.01/0.1
=0.001M
pH= 14 –pOH
=14 –(–(–3))
=14 – 3
=11
4.The stuff that student B mentioned is true. pH meter is measured the pH value of the solution when the reaction of copper ions with hydroxide ions will decrease the hydroxide ion concentration in the solution. However, we can still use the pH meter to monitor the reaction since the water ion-product is a constant. That means the concentration of hydrogen ion times that of hydroxide ion is a constant. The decrease of hydroxide ion concentration means a increase of the concentration of hydrogen ions and so we can use the pH meter to monitor the progress of the reaction.
5.Molar mass of CH3COOH=60
Moles of NaOH used =0.1 x 0.0162
=0.00162
Molarity of CH3COOH in vinegar=0.00162/0.010 x 5
=0.81M
% weight of CH3COOH in vinegar (assume density of vinegar is 1g/mL)
=(0.81 x 60)/1000 x 100%
=4.86%
6.2HCl + Mg(OH)2→ Mg2+ + 2Cl- + 2H2O
2 mole 1 mole
Molar mass of Mg(OH)2=24 + 17 x 2
=58
Moles of Mg(OH)2 in a tablet=0.3/58
=0.00517
Volume of gastric juice that can be neutralized by a tablet of Milk of Magnesium = (0.00517 x 2) / 0.1
=0.1034 L
=103.4mL
7.(a)CaCO3 (s) + 2HCl (aq) → Ca2+(aq) + 2Cl- (aq) + H2O (l) +CO2(g)
(b)
Moles of excess HCl remained=0.1 x 0.05=0.005
Moles of HCl reacted=0.5 x 0.04 – 0.005
=0.015
Weight of CaCO3 in sample=0.015/2x100
=0.75g
% weight of CaCO3 in eggshell sample=0.75/0.8 x100%
=87.9%