25 cm\(^3\) of 0.02 M KOH neutralized 0.03 g of a monobasic organic acid having the general formula C\(_n\)H\(_{2n + 1}\)COOH. The molecular formula of the acid is?
(C = 12, H = 1, O = 16)
Concentration of KOH = 39 x 0.02 = 0.78 g/dm\(^3\)
25cm\(^3\) required 0.03 g of C\(_n\)H\(_{2n + 1}\)COOH
1000cm\(^3\) will require X
\(\frac{{1000}\times{0.03}}{25}\) = 1.2g
C\(_n\)H\(_{2n + 1}\)COOH + KOH → C\(_n\)H\(_{2n + 1}\)COOK + H\(_2\)O
Recall that \(\frac{M_A V_A}{M_B V_B}\) = \(\frac{1}{1}\)
\(\frac{{M_A}\times{1.2}}{{0.02}\times{0.78}}\) = \(\frac{1}{1}\)
M\(_A\) = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
C\(_n\)H\(_{2n + 1}\)COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 60
14n = 60 - 46
n =\(\frac{14}{14}\) = 1
Substituting n = 1 into C\(_n\)H\(_{2n + 1}\)COOH
We'll have CH\(_3\)COOH
∴ CH\(_3\)COOH
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