How many moles of naoh were used in the titration

Asked by admin @ 28/10/2021 in Chemistry viewed by 364 People

Titration of 25.00 ml of an unknown hcl solution with 0.1550 m naoh starts at a buret reading for naoh of 0.33 ml. the phenolphthalein indicator turns light pink in the acid solution for over 30 seconds at a buret reading of 24.19 ml. 2) what was the volume of naoh dispensed? 3) how many moles of naoh were dispensed? 4) write the balanced molecular equation for the neutralization reaction:

Answered by admin @ 28/10/2021

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)

How many moles of naoh were used in the titration

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