Strong acid-Strong base Titration What is the pH of the solution after 9.9 mL of
ID: 504131 • 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
Q1.
Find pH for a solution
V NaOH = 9.9 mL
M = 0.1 M of NaOH
mmol of base (NaOH) = M*V = 9.9*0.1 = 0.99 mmol of NaOH
then
mmol of HCl = MV = (10^-3)(1) = 10^-3 mol = (10^-3)(10^3) = 1 mmol of HCl
the reaction taking place
NaOH + HCl = H2O + NaCl
mmol of HCl left = 1-0.99 = 0.01 mmol of HCl
V total = 1L + 9.9 mL = 1000+9.9 = 1009.9 mL
so
[H+] = mmol/V = 0.01/ 1009.9 = 0.00000990197 M
pH = -log(0.00000990197) = 5.004
Please follow Chegg's Guidelines for posting questions in Q&A. We are only allowed to answer to ONE question per set of Q&A. Post all other multiple questions in multiple set of Q&A