1. III.Neutralization Calculations: 5.0mL of 0.10M NaOH is mixed with 5.0mL of 0.50M HCl. eg: a)Write the net ionic equation for this neutralization reaction. HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) net ionic: H + (aq) + OH - (aq)  H 2 O (l) or: H 3 O + (aq) + OH - (aq)  2 H 2 O (l) b)Calculate the net concentration of H 3 O +. (not all the H 3 O + is neutralized) H 3 O + (aq) + OH - (aq)  2 H 2 O (l) 0.0025mol 0.0005mol There is an excess of H 3 O + of 0.0020 mol.  net [ H 3 O + ] = 0.0020 mol = 0.20M 0.010 L

2. c)Calculate the pH of the final solution. pH = - log [ H 3 O + ] (pOH = - log [ OH - ])  pH = - log (0.20) pH = - (-0.699) pH = 0.70 Note: Where [ H 3 O + ] = 10 x log [ H 3 O + ] = x eg: If [ H 3 O + ] = 1 x 10 -5 then log [ H 3 O + ] = -5 And - log [ H 3 O + ] = 5 = pH

3. pH = - log [ H 3 O + ] -pH = log [ H 3 O + ] antilog (-pH) = [ H 3 O + ]  [ H 3 O + ] = antilog (-2.54) Calculate [H 3 O + ] in a solution that has pH 2.54.eg: [ H 3 O + ] = 2.9 x 10 -3 M

4. Note: If [ H + ] > [ OH - ] Acidic [ H + ] < [ OH - ] Basic [ H + ] = [ OH - ] Neutral @ 25  C If pH < 7 Acidic pH = 7 Neutral pH > 7 Basic pOH < 7 Basic pOH = 7 Neutral pOH > 7 Acidic And: pH + pOH = 14.00

5. Note: for pH and pOH values only digits after the decimal are significant. pH = 8.60 [H3O+] =[H3O+] = 2.5 x 10 -9 M pH = 10.000 [H3O+] =[H3O+] = pH = 7.0 [H3O+] =[H3O+] = pH = 2.521 [H3O+] =[H3O+] = 1.00 x 10 -10 M 1 x 10 -7 M 3.01 x 10 -3 M (2 sig. fig.) (3 sig. fig.) (1 sig. fig.) (3 sig. fig.)