Homework Help: Chemsitry

Posted by someone on Friday, April 27, 2012 at 9:07pm.

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 - DrBob222, Friday, April 27, 2012 at 9:33pm
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)

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