KChapter 16 Assignment t Titration of Strong Acid with Strong Base 4 of 6 100. m

Question

KChapter 16 Assignment t Titration of Strong Acid with Strong Base 4 of 6 100. mL of 0.200 M HCl is titrated with 0.250 M NaOH A titration involves adding a reactant of known quantity to a solution of an another reactant while monitoring the equilibrium concentrations. This allows one to determine the concentration of the second reactant. A pH titration curve specifically monitors the pH as a function of the titrant. PartA What is the pH of the solution after 50.0 mL of base has been added? Express the pH numerically When conducting calculations involving a titration the first step is to write the balanced chemical equation. Then, use the stoichiometric ratios developed from this equation to determine how many moles of each reagent are reacting View Available Hint(s) Submit Part B What is the pH of the solution at the equivalence point? Express the pH numerically View Available Hint(s) Submit

Explanation / Answer

HCl + NaOH -----------> NaCl + H2O

millimoles of HCl = 100 x 0.20 = 20.0

A) millimoles of NaOH added = 50 x 0.25 = 12.5

20 - 12.5 = 7.50 millimoles HCl left

[HCl] = 7.50 / 150.0 = 0.05 M

as HCl is strong acid

[HCl] = [H+] = 0.05 M

pH = - log [0.05]

pH = 1.30

B) as it is strong acid strong base at equivalence point solution will be neutral

pH = 7.00

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