a and b no.
You have 71.7 x 0.182M NaOH = 13.049 mmoles NaOH.
a. At the initial point, (zero mL HCl) you have 0.182M NaOH. So pOH = -log(OH^-) = -log(0.182) = -0.74 and 14.0=-0.74 = 13.26.
b. You add 10mL x 0.2086M = 2.086 mmoles HCl.
..............NaOH + HCl ==>NaCl + H2O
initial mmoles..13.049..0....0......0
add HCl.............2.086.........
change......-2.086.-2.086..2.086.2.086
equil.......10.963.....0....2.086..2.086
pH is determined by the NaOH since that is in excess. (NaOH) = mmoles/mL = 10.963/(71.7+10)mL = 0.134M
Then pOH = 0.872 and pH = 14.0-0.872 = 13.13.
Etc.
3 is determined by excess NaOH
4 is determined by excess NaOH
5 is determined by excess HCl.
Okay so I did this problem, but it seems wrong? I put the answers I got...
Determine the pH of the following tiration at each of the points indicated. A 71.7 mL solution of 0.182 M NaOH is titrated with 0.2086 M HCL.
1.) initial pH= is it 7.86 or 8.51?
2.) after addition of 10.0 mL of HCL= is it 9.98?
3.) after addition of 25.0 mL of HCL= is it 9.50
4.) 50.0 mL of HCL= is it 8.65?
5.) 100.0 mL HCL= is it 5.18?
2 answers
so all of them are wrong XD all right, thanks for the info! I'm going to redo the problem right now
2 answers