An analytical chemist performs a titration to determine the concentration of a barium hydroxide solution, Ba(OH)2\text{Ba(OH)}_2Ba(OH)2, using a 0.100 mol/dm30.100\text{ mol/dm}^30.100 mol/dm3 solution of nitric acid, HNO3\text{HNO}_3HNO3.
In each titration, the chemist pipettes 20.0 cm320.0\text{ cm}^320.0 cm3 of the barium hydroxide solution into a conical flask.
The table shows the volume of nitric acid added in four titration runs:
Titration run1234Volume of acid added (cm3)18.7017.9018.1018.00\begin{array}{|l|c|c|c|c|} \hline \text{Titration run} & 1 & 2 & 3 & 4 \\ \hline \text{Volume of acid added (cm}^3\text{)} & 18.70 & 17.90 & 18.10 & 18.00 \\ \hline \end{array}Titration runVolume of acid added (cm3)118.70217.90318.10418.00Concordant results are those within 0.20 cm30.20\text{ cm}^30.20 cm3 of each other. Identify the concordant results and use them to calculate the average (mean) volume of nitric acid added.
The equation for the reaction is:
2HNO3+Ba(OH)2→Ba(NO3)2+2H2O 2\text{HNO}_3 + \text{Ba(OH)}_2 \rightarrow \text{Ba(NO}_3)_2 + 2\text{H}_2\text{O} 2HNO3+Ba(OH)2→Ba(NO3)2+2H2OCalculate the concentration, in mol/dm3\text{mol/dm}^3mol/dm3, of the barium hydroxide solution.
