1. Consider the titration of 50.00 mL of 0.1000 M HNO2 (Ka = 7.1 x 10-4) with 0.1000 M NaOH. a. Write the equilibrium equation that describes the analyte flask before any base is added. b. Write the reaction that takes place in the analyte flask following the addition of base. c. Use Le Chatelier’s principle to predict how the initial equilibrium, and pH, will change following the addition of base. d. Calculate the pH after adding 0.00, 5.00, 20.00 and 50.00 mL of 0.100 M NaOH. e. Sketch a detailed titration curve based on your calculations from part d. Use a dotted line to predict the pH beyond 50 mL of titrant added. Label the equivalence point.
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The equilibrium equation describing the analyte flask before any base is added is: HNO2(g) + NaOH(aq) → NO2(g) + H2O(l) b. The reaction that takes place in the analyte flask following the addition of base is: Show more…
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1) Calculate the pH of the resulting solution when 100.0 mL of 0.2 M HNO2 is titrated by 0.1 M NaOH after a) 0 mL of NaOH have been added. b) 100 mL of NaOH have been added. c) 150 mL of NaOH have been added. c) 200 mL of NaOH have been added. d) 300 mL of NaOH have been added. The Ka of HNO2 is 4.5 x 10-4
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- Calculate the pH during the titration of $100.0 \mathrm{~mL}$ of $0.230 M$ hydrofluoric acid with $0.500 M \mathrm{NaOH}$ after the addition of $0 \%, 50 \%, 95 \%, 100 \%,$ and $105 \%$ of the base needed to reach the equivalence point. Graph the titration curve (pH vs. volume of $\mathrm{NaOH}$ ), and label the four regions of importance.
Sketch a titration curve (pH versus mL of titrant) for each of the following three hypothetical weak acids when titrated with $0.100 \mathrm{M} \mathrm{NaOH} .$ Select suitable indicators for the titrations from Figure $17-7$. [Hint: Select a few key points at which to estimate the pH of the solution.] (a) $10.00 \mathrm{mL}$ of $0.100 \mathrm{M} \mathrm{HX} ; K_{\mathrm{a}}=7.0 \times 10^{-3}$ (b) $10.00 \mathrm{mL}$ of $0.100 \mathrm{M} \mathrm{HY} ; K_{\mathrm{a}}=3.0 \times 10^{-4}$ (c) $10.00 \mathrm{mL}$ of $0.100 \mathrm{M} \mathrm{HZ} ; K_{\mathrm{a}}=2.0 \times 10^{-8}$
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