Strong acid-Strong base Titration What is the pH of the solution after 9.9 mL of
ID: 504132 • Letter: S
Question
Strong acid-Strong base Titration What is the pH of the solution after 9.9 mL of 0.1 M NaOH is added to 1.0 L of 10^-3 M HCl solution? Assume T = 25 degree C and ionic strength is negligible. Weak acid-Strong base Titration (Use Approximation equations) For a 0.001-M H_2S solution (PKa of H_2S = 7.0), titrated with a strong NaOH solution, determine the following: (a) Initial pH (b) Midpoint pH (c) Equivalence point pH Weak base-Strong acid titration(Use Approximation equations) For a 10^-4-M CH_3COONa (sodium propionate) solution (PKa of CH_3COONa = 4.9), titrated with a H_2SO_4 solution, determine the following: (a) Initial pH (b) Midpoint pH (c) Equivalence point pH Weak acid pH during titration with strong base: Beginning of titration pH 1/2 (pKa + pC) Midpoint of titration pH pKa Equivalence point of titration pH 1/2 (pKa + pKw - pC) Weak base pH during titration with strong acid: Beginning of titration pH pKw - 1/2 (pKb + pC) Midpoint of titration pH pKw - pKb Equivalence point of titration pH 1/2 (pKw - pKb + pC)Explanation / Answer
1. 9.9 mL of 0.1 M NaOH
1.0 L of 10^-3 M HCl solution.
Calculate moles of NaOH and HCl:
Moles of NaOH = Molarity * volume
= 0.10 M * 9.9 mL = 0.99 mmol
Moles of HCl = 10^-3 M * 1.0 L or 1000 mL = 1 mmol
So moles of HCl in excess = 1 - 0.99 = 0.01 mmol
Total volume = 1000 mL + 9.9 mL = 1009.9 mL
[HCl] = [H+] = Moles / volume
= 0.01 mmol / 1009.9 mL = 9.90 * 10^-6 M
pH = - log [H+] = - log (9.90 * 10 ^-6 ) = 5.00