Chemistry Labs

Problem 4

A volume of 31.7 cm3^3 of a 0.1-normal NaOH solution is required for the neutralisation of 0.19 g of an organic acid whose vapour is thirty times as dense as gaseous hydrogen. Give the name and structural formula of the acid. (The acid concerned is a common organic acid.)
Step 1 of 3: Moles of base used
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n(NaOH)=cV=0.1×0.0317=3.17×10−3 moln(\ce{NaOH}) = cV = 0.1 \times 0.0317 = 3.17 \times 10^{-3}\ \text{mol}
Analysis

The moles of NaOH equal the moles of ionisable protons: n(acid)=3.17×10−3/vn(\text{acid}) = 3.17 \times 10^{-3}/v mol, where vv is the basicity of the acid.