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InterviewSolution
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Calculate the normality of 9.0 gram of oxalic acid, `(COOH)_(2)`, dissolved in 250 mL of solution. Strategy: Find the equivalent mass of oxalic acid by dividing its formula mass by its basicity, which is two, because oxalic acid is a dibasic acid. Use the numerical value of equivalent mass to get gram equivalent mass. Then convert grams of `(COOH)_(2)` to equivalent of `(COOH)_(2)`, which lets to calculate the nirmality |
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Answer» Each molecule of oxalic acid contains two acidic hydrogen atoms: `H-O-overset(O)overset(||)(C)-overset(O)overset(||)(C)-O-H` Formula mass of `(COOH)_(2)=90u` Basicity of `(COOH)_(2)=2` Thus, Equivalent mass of oxalic acid `=("Formula mass of" (COOH)_(2))/("Basicity of" (COOH)_(2))` `=(90u)/2=45 u` Hence, gram equivalent mass of oxalic acid is `45 g eq.^(-1)`. Number of equivalents of oxalic acid `=("Mass of"(COOH)_(2))/("Gram equivalent mass of" (COOH)_(2))` `=(9 g)/(45 g eq.^(-1))` `=0.2 eq`. `"Normality" (N)=(no . eq. (COOH)_(2))/V_(mL)xx(1000 mL)/L` `=(0.2 eq.)/(250 mL)xx(1000 mL)/L` `=0.8 eq. L^(-1)` or `0.8 N` |
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