Post a New Question

Chemsitry

posted by .

Hi I have done this problem like 15 times
when I find the mole I get .8 for HCL and .23 for Naoh and the equilibrium conc i get is negative. then I plug it in and get it wrong. Can someone please help me?
Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What is the pH of the solution created by combining 2.30 mL of the 0.10 M NaOH(aq) with 8.00 mL of the
0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)

  • Chemsitry -

    2.30 mL x 0.10 M NaOH = 0.230 millimols.
    8.00 mL x 0.10 M HCl n= 0.800 mmols.
    There is no table to complete so I will make my own.
    .........HCl + NaOH ==> H2O + NaCl
    initial..0.8....0.......0.......0
    add............0.23..................
    change..-0.23..-0.23.....0.23.....0.23
    equil...0.57.....0.......0.23.....0.23
    So you have 0.57 millimoles HCl, no NaOH, and 0.23 millimols NaCl(which does not hydrolyze) in 2.30+8.00 = 10.30 mL.
    (HCl) = mmols/mL = 0.57/10.30 = 0.0553 and convert that to pH.
    ----------------------------------------
    Problem 2.
    2.30 mL x 0.10 M NaOH = 0.230 millimoles
    8.00 mL x 0.10 M HAc = 0.800 mmols.

    ...........HAc + NaOH ==> H2O + NaAc
    initial...0.8.....0........0......0
    add..............0.23
    change...-0.23....-0.23....0.23..0.23
    equil.....0.57.....0......0.23...0.23
    So you have 0.57 mmols HAc, no NaOH, and 0.23 mmoles NaAc so you have a buffered solution. Use the Henderson-Hasselbalch equation if you are familiar with that.
    (HAc) = 0.57/10.23 = ?
    (NaAc) = 0.23/10.23 = ?
    pH = pKa + log (base)/(acid)
    pH = pKa + log (NaAc)/(HAc)

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question