Respuesta :
Q1)
NaOH solution is in the burette and called the titrant
HCl a known volume is in the titration flask and called the titrand.
initial burette reading of NaOH - 0.33 mL
final buretter reading of NaOH - 24.19 mL
therefore volume of NaOH added / dispensed - 24.19 - 0.33 = 23.86 mL
23.86 mL of NaOH was dispensed
Q2)
phenolphthalein is an acid base reaction indicator that shows a colour change from colourless in neutral and acidic media to pink colour in basic media.
phenolphthalein when added to the acid solution is colourless.
Once NaOH is added, OH⁻ reacts with H⁺ in acid and is neutralised. After all the H⁺ in medium is used up, the added OH⁻ is no longer neutralised.
in the presence of excess OH⁻, solution becomes basic and phenolphthalein becomes pink.
number of moles of NaOH = concentration x volume
number of NaOH moles = 0.1550 mol/L x (23.86 x 10⁻³ L) = 3.698 x 10⁻³ mol
NaOH moles added = 3.698 x 10⁻³ mol
Q3)
Neutralisation reaction is the reaction between an acid and base.
H⁺ ions of acid and OH⁻ ions of base react to form water and become neutralised.
when acid and base react they form water and salt.
the reaction is as follows;
HCl (aq) + NaOH (aq) --> NaCl (aq) + H₂O (l)
NaOH solution is in the burette and called the titrant
HCl a known volume is in the titration flask and called the titrand.
initial burette reading of NaOH - 0.33 mL
final buretter reading of NaOH - 24.19 mL
therefore volume of NaOH added / dispensed - 24.19 - 0.33 = 23.86 mL
23.86 mL of NaOH was dispensed
Q2)
phenolphthalein is an acid base reaction indicator that shows a colour change from colourless in neutral and acidic media to pink colour in basic media.
phenolphthalein when added to the acid solution is colourless.
Once NaOH is added, OH⁻ reacts with H⁺ in acid and is neutralised. After all the H⁺ in medium is used up, the added OH⁻ is no longer neutralised.
in the presence of excess OH⁻, solution becomes basic and phenolphthalein becomes pink.
number of moles of NaOH = concentration x volume
number of NaOH moles = 0.1550 mol/L x (23.86 x 10⁻³ L) = 3.698 x 10⁻³ mol
NaOH moles added = 3.698 x 10⁻³ mol
Q3)
Neutralisation reaction is the reaction between an acid and base.
H⁺ ions of acid and OH⁻ ions of base react to form water and become neutralised.
when acid and base react they form water and salt.
the reaction is as follows;
HCl (aq) + NaOH (aq) --> NaCl (aq) + H₂O (l)