Question

an analytical chemist weighs out 0.345 g of an unknown diprotic acid into a 250 ml volumetric flask and dilutes to the mark with distilled water. she then titrates this solution with 0.1800 m naoh solution. when the titration reaches the equivalence point, the chemist finds she has added 23.1 ml of naoh solution.
calculate the molar mass of the unknown acid. round your answer to 3 significant digits.

Explanation:

Step1: Calculate the moles of NaOH

Use the formula \(n = C\times V\).
Given \(C = 0.1800\space M\) and \(V=23.1\space mL=23.1\times10^{- 3}\space L\).
\(n_{NaOH}=0.1800\space mol/L\times23.1\times 10^{-3}\space L = 4.158\times10^{-3}\space mol\)

Step2: Relate moles of NaOH to moles of diprotic acid

For a diprotic acid \(H_2A\) reacting with \(NaOH\) (\(H_2A + 2NaOH=Na_2A + 2H_2O\)), the mole ratio \(n_{H_2A}:n_{NaOH}=1:2\).
So \(n_{H_2A}=\frac{n_{NaOH}}{2}\)
\(n_{H_2A}=\frac{4.158\times 10^{-3}\space mol}{2}=2.079\times10^{-3}\space mol\)

Step3: Calculate the molar mass of the diprotic acid

Use the formula \(M=\frac{m}{n}\).
Given \(m = 0.345\space g\) and \(n = 2.079\times10^{-3}\space mol\)
\(M=\frac{0.345\space g}{2.079\times10^{-3}\space mol}\approx166\space g/mol\)

Answer:

\(166\space g/mol\)