Formula for phthalic acid: HOOC(C₆H₄)COOH
HOOC(C₆H₄)COOH(aq)+2NaOH(aq)
→(OOC(C₆H₄)COO)Na₂(aq)+2H₂O(l)
We shall first determine the number of moles of phthalic acid
used.
Relative Molecular Mass(RMM) of phthalic acid
= 6(RAM of H)+4(RAM of O)+8(RAM of C)
= 6(1)+4(16)+8(12)
= 166
166g= 1 mole of phthalic acid
0.900g= 0.900/166
= 5.42x10⁻ ³ moles of phthalic acid
According to the equation, the ratio of the number of moles of
phthalic acid needed to that of NaOH needed is 1:2. Therefore, the
number of moles of NaOH needed
= 2(number of moles of phthalic acid)
= 2(5.42x10⁻ ³)
= 0.0108
[NaOH] needed
= number of moles of NaOH/volume of NaOH solution in L
= 0.0108/(40.0x10⁻ ³)
= 0.271M
Number of moles of NaOH in 25.0mL solution of 0.271M NaOH
= ([NaOH])x(volume of the solution in L)
= (0.271)x(25.0x10⁻ ³)
= 6.777x10⁻ ³
Formula for benzoic acid: C₆H₅COOH
C₆H₅COOH(aq)+NaOH(aq)
→C₆H₅COONa(aq)+H₂O(l)
According to the equation, the ratio of the number of moles of
benzoic acid needed to that of NaOH needed is 1:1. Therefore, the
number of moles of C₆H₅COOH needed
= number of moles of NaOH needed
= 6.777x10⁻ ³
Relative Molecular Mass(RMM) of benzoic acid
= 7(RAM of C)+6(RAM of H)+2(RAM of O)
= 7(12)+6(1)+2(16)
= 122
1 mole of benzoic acid= 122g
6.777x10⁻ ³ moles of benzoic acid
= (122)(6.777x10⁻ ³)
= 0.827g
Thus, 0.827g of benzoic acid are required.
A 0.900 g sample of phthalic acid re- quired 40.0 mL of NaOH solution for complete...
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