Sodium hydroxide:
sodium hydroxide is very corrosive and can
- Cause severe burns
- Cause permanent eye damage
- Be very harmful if ingested
Hydrochloric Acid (HC l):
HCl is very toxic and can cause following damage:
- Concentrated HCl can form mists. Both the mists are solution have corrosive effect on human tissues.
- Can cause damage your respiratory organs
- Burn your skin severely
- Cause damage to your eye sight if came into contact with eyes.
- It can also harm intestines if drunk.
Ethonoic Acid:
Ethanoic acid has very strong and sharp odour and smell.
Procedure:
Titration of hydrochloric acid with standard sodium hydroxide solution
- Using a pipette and filler, transferred 50.0 cm of NaOH solution into the polystyrene cup. Allowed to stand for a view minutes.
- Recorded the temperature of the solution.
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From a burette, added 5.0 cm3 of HCl solution to the cup.
- Stirred the mixture with the thermometer and recorded its temperature.
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Added successive 5.0 cm3 portions of HCl solution stirring the mixture and recording its temperature after each addition.
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Recorded my results in a copy of result table 14a. Stopped after addition of 50.0 cm3 of acid.
Titration of ethanoic acid with standard sodium hydroxide solution
- Followed the same procedure as I did for the titration of HCl, except that I used ethanoic acid in the burette. When filling the burette, I had to remember to use to correct rinsing procedures.
- Recorded my results in a copy of results table 14b
Results Table 14a, Titrations of hydrochloric acid.
Results Table 14b Titrations of ethanoic acid.
Calculations
Calculate the concentration of each of the acid
In neutralization:
Moles of Base = Moles of Acid
So
No of Moles of NaOH = No of Moles of HCL
Moles
Concentration =
Volume
Concentration of NaOH = 1 mol
Volume of NaOH = 50/1000 = 0.05dm3
Moles = concentration x volume
= 1 x 0.05 = 0.05
SO:
Concentration of HCl = moles/volume
= 0.05/ 0.024
2mol/dm3
Concentration of CH3COOH = moles/volume
= 0.05/0.021
= 2.4mol/dm3
Q4: From the maximum temperature rise, determine the quantity of energy released in each titration. Assume that the specific heat capacity of the solution is the same as water, 4.18 KJKg-1K- .
Ans:
∆ H for HCL:
∆H = MC∆ T
∆H = (volume of NaOH + HCl) x 4.18 x (final temperature- initial temperature)
= (50+21) x 4.18 x (33.5 – 25)
= 0.0 71 x 4.18 x 8.5
= 2.5226 KJ
∆ H for CH3CO2H:
∆H = MC∆ T
∆H = (volume of NaOH + CH3CO2H) x 4.18 x (final temperature- initial temperature)
= (50+19) x 4.18 x (32.5 – 25)
= 0.0 69 x 4.18 x 7.5
= 2.163 KJ
Q5: Calculate the standard enthalpy change of neutralization for each reaction.
Standard enthalpy change for HCl:
∆Ht = 0.071 x 4.18 x 8.5
= 2.5226 KJ
Standard enthalpy change for CH3CO2H:
∆Ht = 0.069 x 4.18 x 7.5
= 2.163 KJ
ANSWERS
- A s long we are considering a strong acid reacting with a strong base , we always dealing with the same reaction;
H+ (aq) + OH- (aq) H2O (l)
This is because strong acids and bases are assumed to be completely ionised, that is why the value is always constant.
-
Hydrochloric acid is less negative than -57 kj mol-1 because:
- Energy was lost to the surrounding because the polystyrene cup was not good insulator.
- Secondly, heat escaped through the hole of the polystyrene cup where the thermometer was inserted to take the temperature reading.
- Weak acids and bases does not fully ionise when they dissolve in water. Ethanoic acid is typical example of a weak acid. It reacts with water to produce hydroxonium and ethonate ions, but the back ward reaction is more successful than the forward one. When ethanoic acid react with water only view percentage of the cid is converted into ions.