Question

an analytical chemist weighs out 0.111 g of an unknown monoprotic acid into a 250 ml volumetric flask and dilutes to the mark with distilled water. he then titrates this solution with 0.0900 m naoh solution. when the titration reaches the equivalence point, the chemist finds he has added 13.1 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 \(C = 0.0900\space M\) and \(V=13.1\space mL=13.1\times10^{- 3}\space L\).
\(n_{NaOH}=0.0900\space mol/L\times13.1\times 10^{-3}\space L = 1.179\times10^{-3}\space mol\)

Step2: Relate moles of acid and NaOH

For a monoprotic acid \(HA\) and \(NaOH\) reaction \(HA + NaOH=NaA + H_2O\), the mole ratio \(n_{HA}:n_{NaOH}=1:1\). So \(n_{HA}=n_{NaOH}=1.179\times 10^{-3}\space mol\)

Step3: Calculate the molar mass of the acid

Use the formula \(M=\frac{m}{n}\). Given \(m = 0.111\space g\) and \(n = 1.179\times10^{-3}\space mol\).
\(M=\frac{0.111\space g}{1.179\times10^{-3}\space mol}\approx94.1\space g/mol\)

Answer:

\(94.1\space g/mol\)