Question 11-3 Titration of Weak Base with Strong Acid; 11-6 Finding the End Point with Indicators You are titrating \( 50.0 \mathrm{~mL} \) of \( 0.200 \mathrm{M} \) sodium propionate \( \left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-} \mathrm{Na}^{+}\right) \)acid with \( 0.100 \mathrm{M} \mathrm{HCl} \). Us the data in the table to answer the questions. \begin{tabular}{|l|l|l|l|} \hline Name & Structure & \( \mathrm{pK} \) & \( \mathrm{K} \) \\ \hline Propanoic acid & \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \) & 4.874 & \( 1.34 \times 10^{-3} \) \\ \hline \end{tabular} a) Assuming that the equivalence point is considered \( 100 \% \) completion, calculate the \( \mathrm{pH} \) at the following points: \( 0 \%, 10 \%, 50 \%, 90 \%, 100 \% \), and \( 120 \% \). b) Sketch the titration curve.
【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/1Sodium propinoate - A- at equivalence point-VHCl = 50 ml x 0.2 M / 0.1 M = 100 mlat 0%-A- + H2O -> HA + OH- 0.2 0 0-x +x +x0.2-x x xKb = x2 / (0.2-x) = 0.75 x 10-9 x = 0.122 x 10-4 MpOH = -log(0.122 x 10-4)pOH = 4.91pH = 14 - 4.91 = 9.09at 10% -Volume of HCl = 10 mlmoles of HCl = 10 ml x 0.1 M = 1 mmolmoles of A- = 50 ml x 0.2 M = 10 mmolA- + HCl -> HA10 1 0-1 -1 +19 0 1pH = pka + log([A^-]/[HA])pH = 4.874 + log(9/1)pH = 5.83at 50% - half equivalence pointpH = pka = 4.874at 90%-Volume of HCl = 90 mlmoles of HCl = 90 ... See the full answer