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vladimir1956 [14]
3 years ago
5

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 phenolpht

halein 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:
Chemistry
1 answer:
Paul [167]3 years ago
8 0
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)
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Now put all the given values in the above formula, we get:

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Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

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