0.0019L mols of NaOH
0.008L x 0.1M HCL = 0.0008 mols HCL
0.00019 - 0.0008 = -0.00061 mols OH left over
Total Volume: 0.0019 + 0.008 = 0.0099
-0.00061 mols/0.0099 L = -0.0616 M [OH-]
pOH = -log [OH-] = -log [-0.0616] = Error
REVISION TO PREVIOUS POST:
What is the pH of the solution created by combining 1.90 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)?
I understood how to do the HC2H3O2, but I can't do the HCl. I am trying to do the same procedure but I am getting a log(-0.00061), which does not exist.Can someone help me please?
3 answers
0.0019L mols of NaOH
That was 0.0019 K x 0.1 M = 0.00019 moles NaOH
0.008L x 0.1M HCL = 0.0008 mols HCL
0.00019 - 0.0008 = -0.00061 mols OH left over
The larger number is 0.0008 so it will be left over. 0.0008 HCl - 0.00019 NaOH = +0.00061 moles HCl left over.
Total Volume: 0.0019 + 0.008 = 0.0099
-0.00061 mols/0.0099 L = -0.0616 M [OH-]
pOH = -log [OH-] = -log [-0.0616] = Error
0.00061 moles HCl/0.0099 = 0.0616 M
pH = -log(0.0616) = 1.21
That was 0.0019 K x 0.1 M = 0.00019 moles NaOH
0.008L x 0.1M HCL = 0.0008 mols HCL
0.00019 - 0.0008 = -0.00061 mols OH left over
The larger number is 0.0008 so it will be left over. 0.0008 HCl - 0.00019 NaOH = +0.00061 moles HCl left over.
Total Volume: 0.0019 + 0.008 = 0.0099
-0.00061 mols/0.0099 L = -0.0616 M [OH-]
pOH = -log [OH-] = -log [-0.0616] = Error
0.00061 moles HCl/0.0099 = 0.0616 M
pH = -log(0.0616) = 1.21
Thank You Dr.Bob. I really appreciate your help.