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Solved: Four solutions of unknown NaOH concentration are titrated with solutions of HCl
Chapter 14, Problem 48P(choose chapter or problem)
Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. Calculate the concentration (in M) of the unknown NaOH solution in each case.
\(\begin{array}{|l|l|l|} \hline \begin{array}{l} \mathrm{NaOH} \text { Volume } \\ (\mathrm{mL}) \end{array} & \begin{array}{l} \mathrm{HCl} \\ \text { Volume } \\ (\mathrm{mL}) \end{array} & {[\mathrm{HCl} \text { (M) }} \\ \hline \text { (a) } 5.00 \mathrm{~mL} & 9.77 \mathrm{~mL} & 0.1599 \mathrm{~M} \\ \hline \text { (b) } 15.00 \mathrm{~mL} & 11.34 \mathrm{~mL} & 0.1311 \mathrm{~M} \\ \hline \text { (c) } 10.00 \mathrm{~mL} & 10.55 \mathrm{~mL} & 0.0889 \mathrm{~M} \\ \hline \text { (d) } 30.00 \mathrm{~mL} & 36.18 \mathrm{~mL} & 0.1021 \mathrm{~M} \\ \hline \end{array}\)
Questions & Answers
QUESTION:
Four solutions of unknown NaOH concentration are titrated with solutions of HCl. The following table lists the volume of each unknown NaOH solution, the volume of HCl solution required to reach the equivalence point, and the concentration of each HCl solution. Calculate the concentration (in M) of the unknown NaOH solution in each case.
\(\begin{array}{|l|l|l|} \hline \begin{array}{l} \mathrm{NaOH} \text { Volume } \\ (\mathrm{mL}) \end{array} & \begin{array}{l} \mathrm{HCl} \\ \text { Volume } \\ (\mathrm{mL}) \end{array} & {[\mathrm{HCl} \text { (M) }} \\ \hline \text { (a) } 5.00 \mathrm{~mL} & 9.77 \mathrm{~mL} & 0.1599 \mathrm{~M} \\ \hline \text { (b) } 15.00 \mathrm{~mL} & 11.34 \mathrm{~mL} & 0.1311 \mathrm{~M} \\ \hline \text { (c) } 10.00 \mathrm{~mL} & 10.55 \mathrm{~mL} & 0.0889 \mathrm{~M} \\ \hline \text { (d) } 30.00 \mathrm{~mL} & 36.18 \mathrm{~mL} & 0.1021 \mathrm{~M} \\ \hline \end{array}\)
ANSWER:Step 1 of 6
The titration of acid-alkali is mainly used for analysing the unknown concentration/potential of a particular alkali or acid. The acid and alkali get neutralized to analyse the unknown potential.