General

[quote=rabbithole]Write out the formula to see how many moles of OH and H you get.

NaOH + HCl --> NaCl + H2O

Since Molarity is moles / L, find out how many moles of reactants there are.

(0.1 mol NaOH / 1 L NaOH) (0.3 L NaOH) = 0.03 mol NaOH
(O.1 mol HCl / 1 L HCl) (0.2 L HCl)= 0.02 mol HCl

This means the 0.03 mol NaOH will completely neutralize 0.02 mol HCl but since more NaOH is present, there'll still be 0.01 mol NaOH left in the solution (0.03 mol NaOH - 0.02 HCl = 0.01 mol NaOH remaining). Remember you mixed the two reactants together so 300 mL + 200 mL = 500 mL or 0.5 L solution.

0.01 mol NaOH remaining / 0.5 L solution = 0.02 M OH.
pOH = -log[OH-] = -log (0.02 M) = 1.70
pH + pOH = 14
14 - 1.70 = 12.3

That's your pH, which makes sense because you added more base than you did acid so it should be a basic solution.

Edit: Ninja'd But she got the same answer too
@CrayonScribble he* LOL sorry. Scrolled too fast. Thought I read a girl's basilID for some reason.[/quote]

Thanks for helping me too! You still gave a really good explanation for the answer.

[quote=rabbithole]Write out the formula to see how many moles of OH and H you get.

NaOH + HCl --> NaCl + H2O

Since Molarity is moles / L, find out how many moles of reactants there are.

(0.1 mol NaOH / 1 L NaOH) (0.3 L NaOH) = 0.03 mol NaOH
(O.1 mol HCl / 1 L HCl) (0.2 L HCl)= 0.02 mol HCl

This means the 0.03 mol NaOH will completely neutralize 0.02 mol HCl but since more NaOH is present, there'll still be 0.01 mol NaOH left in the solution (0.03 mol NaOH - 0.02 HCl = 0.01 mol NaOH remaining). Remember you mixed the two reactants together so 300 mL + 200 mL = 500 mL or 0.5 L solution.

0.01 mol NaOH remaining / 0.5 L solution = 0.02 M OH.
pOH = -log[OH-] = -log (0.02 M) = 1.70
pH + pOH = 14
14 - 1.70 = 12.3

That's your pH, which makes sense because you added more base than you did acid so it should be a basic solution.

Edit: Ninja'd But she got the same answer too [/quote]

Buddy pls, I'mma guy ): But it's good you got the same answer as me because I honestly can't remember the chem I learnt in first yr undergrad too well anymore xd

Write out the formula to see how many moles of OH and H you get.

NaOH + HCl --> NaCl + H2O

Since Molarity is moles / L, find out how many moles of reactants there are.

(0.1 mol NaOH / 1 L NaOH) (0.3 L NaOH) = 0.03 mol NaOH
(O.1 mol HCl / 1 L HCl) (0.2 L HCl)= 0.02 mol HCl

This means the 0.03 mol NaOH will completely neutralize 0.02 mol HCl but since more NaOH is present, there'll still be 0.01 mol NaOH left in the solution (0.03 mol NaOH - 0.02 HCl = 0.01 mol NaOH remaining). Remember you mixed the two reactants together so 300 mL + 200 mL = 500 mL or 0.5 L solution.

0.01 mol NaOH remaining / 0.5 L solution = 0.02 M OH.
pOH = -log[OH-] = -log (0.02 M) = 1.70
pH + pOH = 14
14 - 1.70 = 12.3

That's your pH, which makes sense because you added more base than you did acid so it should be a basic solution.

Edit: Ninja'd But she got the same answer too
@CrayonScribble he* LOL sorry. Scrolled too fast. Thought I read a girl's basilID for some reason.

HAHAH I'm doing this at school right now!
But I shall drop it next year...
Chemistry annoying

[quote=CrayonScribble]Those two solutions negate 1:1 so, you end up with 100mL of 0.1M NaOh, in 500mL.

So effectively the NaOH conc is 0.02M.

Since NaOH is a strong base, we take the log of 0.02 and whatnot and add it to 14, and we get 12.3 pH.

I think, I did this stuff as an undergrad. Not quite sure now.[/quote]

Omg, thank you so much! I spent the whole afternoon trying to do it but I kept getting weird answers.

Those two solutions negate 1:1 so, you end up with 100mL of 0.1M NaOh, in 500mL.

So effectively the NaOH conc is 0.02M.

Since NaOH is a strong base, we take the log of 0.02 and whatnot and add it to 14, and we get 12.3 pH.

I think, I did this stuff as an undergrad. Not quite sure now.