chemistry_concentration&equation問題 (FORM 4)

1) H+(aq) + OH-(aq) → H2O(l)


2)no of mole of the acid=25/1000x0.2=0.005mol
no of mole of NAOH=20/1000x0.5=0.01mol
according to the equation, the mole ratio of H+(aq) : OH-(aq)=1:1
so the NAOH is in excess and the acid is the limiting reactant

so the no of mole of OH-(aq)used=0.005mol
the no of mole of hydroxide ions produced from 25cm3
of 0.2M tartaric acid=0.01-0.005=0.005mol

3)let n be the basicity of the acid
no of mole of the acid/no of mole of NAOH=0.005/0.01=1/n
n=2
so the basicity of the acid is 2

1)
H+(aq) + OH-(aq) → H2O(l)

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2)
The question should be 「number of moles of hydrogen ions」, instead of 「number of moles of hydroxide ions」.

No. of moles of NaOH used = MV = 0.5 x (20/1000) = 0.01 mol
Since one mole of NaOH gives one mole of OH- ions.
No. of moles of OH- ions reacted = 0.01 mol

H+(aq) + OH-(aq) → H2O(l)
Mole ratio H+ : OH- = 1 : 1
0.01 mol of OH- ions is reacted.
Hence, no. of moles of H+ ions reacted = 0.01 mol
All H+ ions come from the ionization of tartaric acid.
Hence, no. of moles of H+ ions produced from tartaric acid = 0.01 mol

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3)
No. of moles of tartaric acid = MV = 0.2 x (25/100) = 0.005 mol
No. of moles of H+ ions produced from tartaric acid = 0.01 mol
No. of moles of H+ ions produced by 1 mole of tartaric acid = 0.01/0.005 = 2 mol
Hence, basicity of tartaric acid = 2

1.
C4H6O6 + 2 NaOH --------------- Na2C4H4O6 + 2 H2O

2.
25/1000 x 0.2 x 2 = 0.01 mol

3. 2