Acid- Base Reactions: Titrations help.?

Question

Just started this today, not sure if I'm doing this right.

I have 10 mL of 0.1 M HCL and 20 mL of 0.1 M NaOH. It asks for the moles of H+, Moles of OH-, Moles of excess H+ or OH-, Total Volume, Concentration of {H+} or {OH-}, and finally the PH.

I can get most of the answers, and I get 1 Mol, 2 Mols, 1 Mol, 30 mL Total, and 33.333 for the Molarity, respectively. but my answer for pOH is approx. -1.5 making my pH a very low number... I know negative Phs are possible in calculations, but I'm not sure about this.

Answer 1

0.1M means 0.1 mole in one liter. You have 10mL or 0.01L of solution, it therefore contains 0.001 moles of HCl; 0.001 moles of H+ and 0.001 moles of Cl-. For the NaOH, you have 0.1 M 20mL solution, this has 0.002 moles NaOH; 0.002 moles of OH- and 0.002 moles of Na+. When reacted, 0.001 mole of HCl will react with 0.001 mole of NaOH to produce 0.001 mole of NaCl in a total of 30mL of solution. There will be 0.001 mole of OH- left unreacted in 30mL of solution so the concentration is 0.001/0.03 = 0.033 M OH- The product of [H+] and [OH-] = 10^-14, so [H+] = 3*10^-13, and the pH is -log(3*10^-13) = 12.5

Answer 2

Negative pHs aren't possible, so you've gotten that part wrong.

Your volume answer is correct though

HCl+NaOH ==> NaCl+H2O (2 is supposed to be a subscript)

The pH of the NaCl and H2O solution is 7, because saltwater is a neutral solution

I dont know about the concentration though