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

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)

Strong Acid with Strong BaseThere 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.

Example Problem100 mL of 0.20 M HCl is added to 50 mL of 0.30 M NaOH. What is the pH of the resulting solution?

Example Problem100 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 → H2O + NaCl

Starting moles: 0.020 0.015 0* 0Moles used/made: 0.015 0.015 0.015 0.015Moles after: 0.005 0 0.015 0.015

*There's already water around because these are aqueous, but the amount doesn't really matter.

Example Problem100 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 → H2O + NaCl

Starting moles: 0.020 0.015 0* 0Moles used/made: 0.015 0.015 0.015 0.015Moles 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.

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**

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

Example 1: Excess Acid100 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- → H2O + F-

Starting moles: 0.020 0.015 0 0Moles used/made: 0.015 0.015 0.015 0.015Moles after: 0.005 0 0.015 0.015

Example 1: Excess Acid100 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- → H2O + F-

Starting moles: 0.020 0.015 0 0Moles used/made: 0.015 0.015 0.015 0.015Moles 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.

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 Ka of HF:

HF ⇆ H+ + F-

Initial: 0.033 0 0.10Change: -x +x + xEquilibrium: 0.033-x x 0.10+x

Ka = 7.2x10-4 = (x)(0.10+x)/(0.033-x) x = 0.000235

pH = -log(0.000235) = 3.63

Example 2: Stoichiometric100 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- → H2O + F-

Starting moles: 0.020 0.020 0 0Moles used/made: 0.020 0.020 0.020 0.020Moles after: 0 0 0.020 0.020

Example 2: Stoichiometric100 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- → H2O + F-

Starting moles: 0.020 0.020 0 0Moles used/made: 0.020 0.020 0.020 0.020Moles 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.

[F-] = 0.020 moles/ 0.20 L = 0.10 M

Now set up the ICE box for the Kb of F-:

F- + H2O ⇆ HF + OH-

Initial: 0.10 --- 0 0Change: -x --- +x + xEquilibrium: 0.10-x --- x x

Kb = 1.39x10-11 = (x2)/(0.10-x) x = 1.18x10-6

pOH = -log(0.00000118) = 5.93pH = 8.07

Example 2: Stoichiometric

Example 3: Excess Base100 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- → H2O + F-

Starting moles: 0.020 0.040 0 0Moles used/made: 0.020 0.020 0.020 0.020Moles after: 0 0.020 0.020 0.020

Example 3: Excess Base100 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- → H2O + F-

Starting moles: 0.020 0.040 0 0Moles used/made: 0.020 0.020 0.020 0.020Moles 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.

[OH-] = 0.020 moles/ 0.30 L = 0.0667 M

pOH = -log(0.0667) = 1.18pH = 12.82

Example 3: Excess Base

SummaryStrong with Strong: find leftover amount

Weak with Strong:Strong Limiting: ICE box for weakStoichiometric: ICE box for conjugate of weakExcess Strong: pH or pOH of leftover strong