1. Mixing Acids With Bases (Disclaimer: not even remotely as visually stimulating as the picture below)

2. Most Important: Identify your acid-base combination as: 1. Strong with strong (pretty easy) 2. Strong with weak (a little tougher, but not too bad) 3. Weak with Weak (ugh. We just won't do it)

3. Strong Acid with Strong Base There is no equilibrium. The reaction is: Acid + Base → Neutral Stuff If you have the perfect stoichiometric ratio of acid and base (usually 1:1, but not always), then the pH is 7. If you have extra acid or base, just figure out how much is left over.

4. Example Problem 100 mL of 0.20 M HCl is added to 50 mL of 0.30 M NaOH. What is the pH of the resulting solution?

5. Example Problem 100 mL of 0.20 M HCl is added to 50 mL of 0.30 M NaOH. What is the pH of the resulting solution? HCl + NaOH → H 2 O + NaCl Starting moles: 0.020 0.015 0* 0 Moles used/made: 0.015 0.015 0.015 0.015 Moles after: 0.005 0 0.015 0.015 *There's already water around because these are aqueous, but the amount doesn't really matter.

6. Example Problem 100 mL of 0.20 M HCl is added to 50 mL of 0.30 M NaOH. What is the pH of the resulting solution? HCl + NaOH → H 2 O + NaCl Starting moles: 0.020 0.015 0* 0 Moles used/made: 0.015 0.015 0.015 0.015 Moles after: 0.005 0 0.015 0.015 Water and salt won't affect pH, so just look the [HCl] left over: 0.005 moles / 0.15 L = 0.033 M HCl pH = -log(0.033) = 1.48 *There's already water around because these are aqueous, but the amount doesn't really matter.

7. Weak Acid with Strong Base (The process is the same with weak base and strong acid) Acid + Base → Conjugate base + Neutral thing ** THIS IS NOT AN EQUILIBRIUM**

8. Weak Acid with Strong Base (The process is the same with weak base and strong acid) Acid + Base → Conjugate base + Neutral thing ** THIS IS NOT AN EQUILIBRIUM** Because a strong base is involved, this goes 100%. There will still be equilibrium involved in solving this, but it will be K a, not this reaction. There are three scenarios here: * Excess acid * Perfect stoichiometric ratio * Excess base

9. Example 1: Excess Acid 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 50 mL of 0.30 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.015 0 0 Moles used/made: 0.015 0.015 0.015 0.015 Moles after: 0.005 0 0.015 0.015

10. Example 1: Excess Acid 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 50 mL of 0.30 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.015 0 0 Moles used/made: 0.015 0.015 0.015 0.015 Moles after: 0.005 0 0.015 0.015 There are two big differences here from the previous example: 1. The leftover acid is weak (we will have to do a Ka ICE box) 2. The conjugate base is a weak base (fluoride). We can't ignore it.

11. Example 1: Excess Acid [HF] = 0.005 moles / 0.15 L = 0.033 M [F - ] = 0.015 moles/ 0.15 L = 0.10 M Now set up the ICE box for the K a of HF: HF ⇆ H + + F - Initial: 0.033 0 0.10 Change: -x +x + x Equilibrium: 0.033-x x 0.10+x K a = 7.2x10 -4 = (x)(0.10+x)/(0.033-x)x = 0.000235 pH = -log(0.000235) = 3.63

12. Example 2: Stoichiometric 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 100 mL of 0.20 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.020 0 0 Moles used/made: 0.020 0.020 0.020 0.020 Moles after: 0 0 0.020 0.020

13. Example 2: Stoichiometric 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 100 mL of 0.20 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.020 0 0 Moles used/made: 0.020 0.020 0.020 0.020 Moles after: 0 0 0.020 0.020 When it was strong/strong, this was just neutral. But now: We made a weak base (fluoride). This will be a basic pH.

14. [F - ] = 0.020 moles/ 0.20 L = 0.10 M Now set up the ICE box for the K b of F - : F - + H 2 O ⇆ HF + OH - Initial: 0.10 --- 0 0 Change: -x --- +x + x Equilibrium: 0.10-x --- x x K b = 1.39x10 -11 = (x 2 )/(0.10-x)x = 1.18x10 -6 pOH = -log(0.00000118) = 5.93 pH = 8.07 Example 2: Stoichiometric

15. Example 3: Excess Base 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 200 mL of 0.20 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.040 0 0 Moles used/made: 0.020 0.020 0.020 0.020 Moles after: 0 0.020 0.020 0.020

16. Example 3: Excess Base 100 mL of 0.20 M HF (K a = 7.2x10 -4 ) is combined with 200 mL of 0.20 M NaOH. What is the pH of the resulting solution? HF + OH - → H 2 O + F - Starting moles: 0.020 0.040 0 0 Moles used/made: 0.020 0.020 0.020 0.020 Moles after: 0 0.020 0.020 0.020 Yes, there is some fluoride here too (weak base), but......there's a bunch of strong base left over! Let's just ignore the weak base.

17. [OH - ] = 0.020 moles/ 0.30 L = 0.0667 M pOH = -log(0.0667) = 1.18 pH = 12.82 Example 3: Excess Base

18. Summary Strong with Strong: find leftover amount Weak with Strong: Strong Limiting: ICE box for weak Stoichiometric: ICE box for conjugate of weak Excess Strong: pH or pOH of leftover strong