Question

The flask shown here contains 1.04 g of acid and a few drops of phenolphthalein indicator dissolved in water. The buret contains 0.280 M NaOH. When titrated the end point of the solution is 26 mL.

What volume of base is needed to reach the end point of the titration?

Assuming the acid is monoprotic, what is its molar mass?

Answer 1

we can write here first reaction of acid base titration, (assuming monotropic acid HCl or CH3COOH),

CH3COOH aq + NaOH aq -------------- CH3COO^- Na ⁺ + H2O (BALANCED REACTION),

FROM THE REACTION WE CAN ASY THAT , for the one mole of monotropic acid one mole of base is required to complete reach the end point of reaction (i.e . to neutralise acid ),

we have taken here example of monotropic acid as acetic acid ,

we have given the quantity of acid in gms =1.04 gms ,

the end point of solution is given = 26 ml , (volume after addition of base in acid ),

but the volume of acid solution is not given ,

we can calculate the moles of acid from its molar mass as,

moles= mass in gms/ mol.wt, (assume here acetic acid)

moles= 1.04/60= 0.01733 moles of acetic acid , for neutarlise 0.01733 moles of acetic acid we required 0.01733 moles of NaOH solution , here we have given the molarity of NaOH SOLUTION = 0.280 M,

WE WILL NOW CALCULATE THE MOLES OF NaOH AS,

Moles =molarity* Volume IN L,

MOELS= 0.280*1 L ( VOLUME OF NaOH, assume here we have 1000 ml 0.280 M NaOH Solution)

moles= 0.280 *1=0.280 moles of NaOH FOR 0.280 =1000 ML, THEN FOR 0.01733 MOLES =?,

61.89 mL 0.280 M Naoh solution is required . but we have given the volume after reaction is 26 ml ,

which is 61.89/26=0.42 ,

then which is divided by =60/0.42=142.85 ( molar mass of acid ),

1.04/142.85=0.0072 moles of acid = 0.0072 moles of Naoh,

then 0.280 =1000 ml,

0.0072= ?,

0.0072*1000=7.2/0.280=25.71 ml NaOH ,

MOLAR MASS OF ACID = 142.85