Acids and bases behave differently in water. Acids and bases are categorized into two kinds as strong and weak. pH value is used to express strength of acids and bases. Acids and bases are used to manufacture different compounds (NaOH is used to prepare soap), but Acids and bases are toxic and harmful to people and environmental.
Substances which react with water and release H3O+(aq)( or H+ ). Acids shows a less pH values. Acidic solutions can conduct electricity. Acids react with alkalis and produce salt and water.
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
Acidic substance which dissociate completely into H3O+( or H+ ) and related anion in the water. These anions are very stable in the solution. Cl-, NO3-, ClO4- are some of stable anions in the water.
Acidic substance, dissociate partially into H3O+( or H+ ) and related cation in the water.
Substances which reacts with water and release OH-(aq). Alkalis provide ions to the solution, therefore they conduct the electricity. Bases have negative charges(OH-,NH2-) or lone pair (NH3). Alkalis react with acids and produce salts and water.
BOH(aq) → B+(aq) + OH-(aq)
Bases dissociate completely into hydroxide (OH-) and related cation in the water. These cations are very stable in the water. Some of these cations are Na+, K+. This is happened due to these cations are stabled by ion-dipol forces.
Base substance, which dissociate partially into hydroxide ( OH- ) and related cation in the water.
Acids and bases react and give salt and water as products. If HA is an acid and BOH is a base following reaction occurs.
HA(aq) + BOH(aq) → AB(aq) + H2O(l)
HCl is a strong acid and NaOH is a strong base. They react very easily and violently. This is an
exerthormic reaction. This reaction should be done carefully. Both HCl and NaOH react to 1:1 molar ratio.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Yes. Barium hydroxide ( Ba(OH)2 ) is a strong base. Barium hydroxide dissociate completely into Ba2+ ion and OH- ions in the water. Ba2+ is a stable cation in the water. When bases release a stable cation to the water, those bases are defined as strong bases.
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)
Barium hydroxide (Ba(OH)2(aq)) is a strong base. Sulfuric (H2SO4(aq) ) is a strong acid. So Ba(OH)2(aq) react with HCl and give barium sulphate ( BaSO4(s) ) and water as products. BaSO4 is a white precipitate. When reaction occurs, you can see this white precipitate forms bottom of the plask. This is also an exerthormic reaction.
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)
Aqueous NaOH is a strong base and CH3COOH is a weak acid. They react together and give sodium ethanoate which is a weak base.
These calculations mostly include concentration calculations. When you know concentration of an acid or base and volume of solution, you can calulate amount of moles of solution. In these calculation we use
C = n / V expression most of times. ( C - Concentration , n = number of moles , V= Volume of solution )
You are provided 1moldm-3 NaOH solution. Calculate the concentration of Na+ and OH- given by dissociation of NaOH.
NaOH is a strong base. So it dissociate completely in water and give Na+ ion and OH- ion.
NaOH(aq) → Na+(aq) + OH-(aq)
According to the stoichiometry we see, one NaOH molecule breaks into one Na+ ion and OH- ion. Therefore, there are 1moldm-3 Na+ ion concentration and 1moldm-3 OH- ion concentration.
You are given two solutions 0.02 moldm-3 HCl 50cm3 and 0.01moldm-3 NaOH 100cm3. Both solutions are mixed slowly and let to react with each other. Calculate followings.
Reaction of NaOH and HCl
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
To calculate number of moles of HCl and NaOH we use the relationship of
Concentration = moles / volume of solution
moles = concentration * volume of solution
Amount of moles of HCl = 0.02 moldm-3 * 50*10-3dm3
Amount of moles of HCl = 0.001 mol
Amount of moles of NaOH = 0.01 moldm-3 * 100*10-3dm3
Amount of moles of NaOH = 0.001 mol
In the balanced equation, you can see NaOH and HCl react 1:1 molar ratio. Therefore equal amount of HCl and NaOH react. In each solutions, there are 0.001 mol. So all of HCl and NaOH react and finish. And 0.001 mol of NaCl forms. Now the volume of total solution is 100 cm-3.
Concentration of NaCl = 0.001mol / 100*10-3dm-3
Concentration of NaCl = 0.01moldm-3