Question

2. Calculate the pH of 30.0 mL. of 0.100 M acetic acid mixed with 40.0 mL of 0.100 M sodium acetate. Ma Ac : rno 01-.IM ° LOmL、4f nol NaAc 3. Calculate the pH of 20.0 mL. of the solution prepared in Problem 2 mixed with 20.0 mL of water continued on back) 17-5 16-3

please show all calculations as well

Answer 1

Pka for acetic acid = 4.75 concentration of acetic acid = 0.100 M = 0.100 mol/l volume of acetic acid solution = 30.0 ml = 30.0 l/1000 = 0.03L no of moles of acetic acid = 0.100 mol/l times 0.03L = 0.003 mol concentration of sodium acetate = 0.100M = 0.100 mol/l volume of sodium acetate solution = 40.0 ml = 40.0l/1000 = 0.04 L no of moles of sodium acetate = 0.100 mol/l times 0.04 l = 0.004 mol total volume of solution = (0.03 + 0.04)L = 0.07 L [CH_3 COOH] = 0.003/0.07L = 0.0428 mol/l = 0.0428M [CH_3 COONa] = 0.004 mol/0.07L = 0.0571 mol/l = 0.0571M from Henderson equation PH = Pka + log [conjugate base]/[acid] Ph = 4.75 + log [CH_3 COONa]/[CH_3COOH] = 4.75 + log (0.0571M/0.0428M) = 4.75 + log 1.33