FA 1 is a solution containing 5.00g dm-3 of a dibasic acid HOOC(CH2)nCOOH. FA 2 is a solution containing 5.00g dm-3 of sodium hydroxide. 25.0cm³ of the acid solution FA 1 needed 17.00cm³ of sodium hydroxide solution, FA 2, for neutralisation using phenolpthalein as an indicator.
Determine the value of n in the formula of the acid.
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Answer:
Dibasic acid = 2H+
Therefore, 2 moles of HOOC(CH2)nCOOH will react with 1 mole of NaOH.
concentration of NaOH = 5g /dm³
= 0.125 mol/ dm³
number of moles of NaOH reacted = 17/1000 x 0.125
= 0.002125 mol
no. of moles of HOOC(CH2)nCOOH in 25.0 cm³ = 0.002125 / 2
= 1.0625 x 10-3 mol
no. of moles of in HOOC(CH2)nCOOH 1 dm³ = (1.0625 x 10-3) / 25 x 1000
= 0.0425 mol
Mr of HOOC(CH2)nCOOH = 5/ 0.0425 = 117.6 (1d.p)
1+16+16+12+12n+2n+12+16+16+1 = 117.6
14n = 27.6
n = 1.97 = 2 (nearest whole number)
