Titrating Sodium hydroxide with an unknown molarity, against hydrochloric acid to find its' molarity.

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Titrating Sodium hydroxide with an unknown molarity, against hydrochloric acid to find its' molarity.

P2aP2aAim: - To find out the concentration of sulphuric (VI) acid by performing a titration with sodium carbonate solution.

Introduction

Sodium carbonate is an alkali, meaning that it has basic properties and has a pH that is greater than pH7, which is neutral. All pHs that are lower than pH7 are considered to be acidic. The pH scale runs from pH1, to pH14.

A base is a solution/substance, which receives hydrogen ions (H+, also referred to as a proton) willingly, whereas acids are proton donators, this means that they give hydrogen ions off when in solution. The more basic something is, the more hydrogen ions it can receive and the more acidic something is, the more hydrogen ions it gives out, pH is a measure of this.

In this particular titration, we will be neutralising the acid (H2SO4) with an alkali (Na2CO3), which is a carbonate. When an acid and a carbonate are reacted, the resulting products are a salt, carbon dioxide and water. These equations show what reactants and products are involved in this experiment: excellent shows you are thinking coherently about the problem evidence for P11a

H2SO4 (?q) + Na2CO3 (?q) --> Na2SO4 (?q) + H2O (?) + CO3 (g)

mole + 1 mole --> 1 mole + 1 mole + 1 mole

Hydrochloric acid (?q) + Sodium hydroxide (?q) --> Sodium chloride (?q) + Water (l)

The acid (H2SO4, in this case) will release protons when the base (Na2CO3) is added to it, which will receive them. Once the solution has been neutralised, it will no longer be able to receive any more protons. We can test the pH of solutions by using various different types of indicators. Good chemical knowledge P11a

Methyl orange is the indicator that will be used within this experiment. When methyl orange is present in acidic solutions, it is red; in basic solutions it is yellow. When the concentrations of two solutions are equal, methyl orange is orange; a mixture of the two colours. When methyl orange just changes from orange to red, this is when we know that we have neutralised the solutions, this is the end point. We know the end point of the titration, as the solution within the conical flask will just turn a light pink. The end point of the titration is the visible colour change that can be seen. The calculated end point is known as the stoichiometric point, this is the actual volume of acid, which is used in order to neutralise the alkali. This should be 25cm3 of the acid as we are using this volume of alkali within the conical flask.

In order to find out what the concentration of the sulphuric acid is, I need to have a standard solution. A standard solution is a solution, good P11a that has been made, the concentration of which is known. As I know the concentration of the standard solution, I can work out what the concentration of the sulphuric acid with the unknown concentration is.

A primary standard needs to meet certain specification. They are: good P11a

* It must be available in a highly pure state

* It must be stable in air at ordinary temperatures

* It must be easily soluble in water

* It must have a high molar mass

* It must be in solution when used in volumetric analysis , must undergo complete and rapid reaction

The concentration of the sulphuric acid is said to be between 0.15moldm-3 and 0.05moldm-3. In order to make up the standard solution, I need to estimate the concentration of the sulphuric acid that we are using. To do this I need to take the mean of the concentrations that we are told the acid could be.

0.05moldm-3 + 0.15moldm-3 = 0.10moldm-3

2

good P11 a and the calculation part is P11b

Therefore, I need make up a standard solution of the sodium carbonate solution with a concentration of 0.10moldm-3. I will need to make up this solution with the appropriate concentration of sodium carbonate solution, so that approximately 25cm3 of the base reacts will react with 25cm3 of the sulphuric acid.

To do this I need to calculate the concentration of sulphuric acid, I know that the concentration of the acid is roughly 0.10moldm-3. We must assume that it is this number as we do not know otherwise. P11b

From the equation for this reaction above, I know that 1 mole of sodium carbonate will react with 1 mole of sulphuric acid. There is an equation that links moles, volume and concentration.

moles = volume x concentration

= 0.025dm3 x 0.10moldm3

= 0.0025 moles of sulphuric acid

Therefore 0.025dm3 of sodium carbonate solution is equivalent to 0.0025 moles of sulphuric acid. We know that the sodium carbonate solution needs to be 0.10 moldm-3 in order to neutralise the sulphuric acid.

Therefore to neutralise 25cm3 of 0.10moldm-3 sulphuric acid I need to use 0.0025 moles of sodium carbonate solution, I can now work out the mass of anhydrous sodium carbonate that I will need to use in order to make up the standard solution. There is an equation that links together moles, the mass and the molar mass of the substance.

Moles = mass

molar mass

0.0025 = mass

(2x23+12+3x16)

0.0025 = mass

106g

Mass = 0.0025x106

= 0.265g

Therefore 0.265g of solid anhydrous sodium carbonate would be needed to be diluted with distilled water to produce 0.025dm3 of sodium carbonate solution.

However, we need to make up 0.25dm3 of sodium carbonate solution; this means that in order to find the correct mass of anhydrous sodium carbonate that needs to be used, so I therefore need to multiply the above answer by ten.

Mass = 0.265 x 10

= 2.65g

As the top pan balance that I have access to only reads to 0.01g, I will need to round the mass up so that it will be possible to actually weigh out this amount. This will mean that my results will not be as precise as they should be. However, as this is the most precise weighing equipment there is, I must use it. Good point !
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The reaction between the anhydrous sodium carbonate and the distilled water will be an exothermic one. This is as energy is released as the anhydrous solid is being hydrated by the water molecules; most of this energy is given out in the form of heat energy, which is why it is an exothermic reaction. The energy that is used within this system -24.6 ?H/kJmol-1 ?, as the number of a minus figure, we can tell that the energy is being given out and not taken in.

? Chemical Data Book by JG Stark and HG Wallace. Second ...

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