Please help with question 1! (Please notice it goes from the first page all the
ID: 505403 • Letter: P
Question
Please help with question 1! (Please notice it goes from the first page all the way to the box on the second page!)
(Also, could you please help with questions 2 and 3?!)
Experiment 10: Titration Curve of a Strong Acid and a Strong Base Prelab PracticeAssignment 1. Quentin Quickdigits has 25.00 mL of a sodium hydroxide solution of concentration 0.2500 M. He adds 10.00 mL of a hydrochloric acid solution of concentration 0.2000 M. His goal is to figure out the pH of the sodium hydroxide solution before and after addition of acid sodium hydroxide pH before acid addition Enter the sodium hydroxide pH into the box and show in the space below it how you arrived at your answer pH Sodium hydroxide solution after acid addition Let na be the number of moles of acid in the 10.00 mL HCl sample, and let nb(before) and nb(after) be the number of moles of base in the NaoH sample before and after addition of the acid, respectively. Enter their numerical values in the table and show in the space below it how you got your answers. mol nb (before) na mol nb (after) molExplanation / Answer
1) The balanced chemical equation for the reaction is
NaOH + HCl -----> NaCl + H2O
As per the stoichiometric reaction,
1 mole NaOH = 1 mole HCl
Since NaOH is a base, we must find out pOH first. For that purpose, we need to evaluate [OH-].
NaOH is a strong electrolyte and ionizes completely into Na+ and OH-. Therefore, [OH-] = 0.2500 M.
pOH = -log [OH-] = -log (0.2500) = 0.6020
Therefore, pH = 14 – pOH = 14 – 0.6020 = 13.398 (ans).
Moles of HCl added, na = (volume of HCl in L)*(concentration of HCl) = (10.00 mL)*(1 L/1000 mL)*(0.2000 mole/L) = 0.002 mol (ans).
Moles of NaOH before addition, nb (before addition) = (volume of NaOH in L)*(concentration of NaOH) = (25.00 mL)*(1 L/1000 mL)*(0.2500 mol/L) = 0.00625 mol (ans).
0.002 mole of HCl will neutralize 0.002 mole of NaOH. Therefore, moles of NaOH after addition of acid, nb (after addition) = (0.00625 – 0.002) mol = 0.00425 mol (ans)
Final volume of the solution, V = (volume of NaOH + volume of HCl) = (25.00 mL) + (10.00 mL) = 35.00 mL = (35.00 mL)*(1 L/1000 mL) = 0.035 L (ans).
Molarity of NaOH = moles of unreacted NaOH/volume of solution in L = (0.00425 mol)/(0.035 L) = 0.1214 mol/L = 0.1214 M (ans).
Since molarity of unreacted NaOH = 0.1214 M, therefore, [OH-] = 0.1214 M.
pOH = -log [OH-] = -log (0.1214) = 0.9158 0.916 (ans)
pH = 14 – pOH = 14 – 0.916 = 13.084 (ans).
2) At the equivalence point, moles of acid added = moles of base reacted.
Moles of NaOH reacted = 0.00625.
Therefore, moles of HCl added = 0.00625.
Volume of HCl required to reach the equivalence point = moles of HCl added/concentration of HCl = (0.00625 mol)/(0.2000 mol/L) = 0.03125 L = (0.03125 L)*(1000 mL/1 L) = 31.25 mL (ans).
3) The neutralization of NaOH with HCl produces NaCl and H2O. NaCl is neither acidic nor basic and hence doesn’t contribute to the pH of the solution. H2O is neutral and has a pH of 7.00. Thus, the pH at the equivalence point of a strong acid and strong base titration is expected to be 7.00 (ans).