Hydrochloric acid (HCl) of known concentration was added from a burette to neutralise 25.0 cm³ portions of sodium hydroxide (NaOH) solution of unknown concentration in apparatus A - Leaving Cert Chemistry - Question 2 - 2017

Apparatus A (the conical flask) is rinsed with deionised water or distilled water to remove any impurities that might affect the reaction.

The piece of apparatus B is a wash bottle.

The wash bottle is used for rinsing the conical flask and other equipment, ensuring that no contaminants are present that can react with the solutions.

A suitable indicator for this titration is methyl orange.

Placing apparatus A on a white tile helps to see the color change of the indicator clearly, making it easier to determine the endpoint of the titration.

The advantage of carrying out a rough titration is to obtain an approximate endpoint of the reaction, which allows for more accurate titrations to follow.

The average volume of hydrochloric acid should be calculated as follows:

egin{align*} ext{Average volume} &= rac{19.6 + 19.5}{2} = 19.55 ext{ cm}^3 ext{But since we include the rough titration, we get:} ext{Average volume} &= rac{19.6 + 19.5 + 19.9}{3} = 19.68 ext{ cm}^3 ext{(Rounded for accuracy)} ext{Therefore, using } 19.55 cm³ is sufficient for calculations. egin{align*}

This balanced equation shows the reaction between hydrochloric acid and sodium hydroxide to form sodium chloride and water.

Using the molarity information:

1. Calculate moles of HCl used: ext{moles of HCl} = 0.1 imes rac{19.55}{1000} = 0.001955
2. From the balanced equation, the mole ratio of HCl to NaOH is 1:1. Therefore, moles of NaOH = moles of HCl = 0.001955 moles.
3. Volume of NaOH used = 0.025 L (since 25.0 cm³).
4. Molarity of NaOH: M = rac{ ext{moles}}{ ext{volume}} = rac{0.001955}{0.025} = 0.0782 ext{ mol L}^{-1}
5. To find grams per litre, we use the molar mass of NaOH (approximately 40 g/mol): $0.0782 ext{ mol L}^{-1} imes 40 ext{ g/mol} = 3.13 ext{ g L}^{-1}$