Go to a listing of many types of acid base problems and their solutions

**Some discussion.**

Important note: all constants refered to: K_{c}, K_{w}, K_{a}, and K_{b} are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.

Basic Information

(1) Weak bases are less than 100% ionized in solution.

(2) Ammonia (formula = NH_{3}) is the most common weak base example used by instructors.

The following equation describes the reaction between ammonia and water:

NH_{3}+ H_{2}O ⇌ NH_{4}^{+}+ OH¯

Note that it is a reaction that comes to a state of equilibrium.

The equilibrium constant for this reaction is written as follows:

[NH _{4}^{+}] [OH¯]K _{c}=––––––––––– [NH _{3}] [H_{2}O]

However, in pure liquid water, [H_{2}O] is a constant value. To demonstrate this, consider 1000 mL of water with a density of 1.00 g/mL. This 1.00 liter (1000 mL) would weigh 1000 grams. This mass divided by the molecular weight of water (18.0152 g/mol) gives 55.5 moles. The "molarity" of this water would then be 55.5 mol / 1.00 liter or 55.5 M.

The solutions studied in introductory chemistry are so dilute that the "concentration" of water is unaffected. So 55.5 molar can be considered to be a constant if the solution is dilute enough.

Moving [H_{2}O] to the other side gives:

[NH _{4}^{+}] [OH¯]K _{c}[H_{2}O] =––––––––––– [NH _{3}]

Since the term K_{c} [H_{2}O] is a constant, let it be symbolized by K_{b}, giving:

[NH _{4}^{+}] [OH¯]K _{b}=––––––––––– [NH _{3}]

This constant, K_{b}, is called the base ionization constant. It can be determined by experiment and each base has its own unique value. For example, ammonia's value is 1.77 x 10¯^{5}.

From the chemical equation above, it can be seen that NH_{4}^{+} and OH¯ concentrations are in the molar ratio of one-to-one. This will have an important consequence as we move into solving weak acid poblems.

**Example #1:** What is the pH of a 0.100 M solution of ammonia? K_{b} = 1.77 x 10¯^{5}

Some facts of importance before the solution:

(1) You know this is a weak base for two reasons:(a) You memorized a short list of strong bases. (You did, didn't you?) Everything else is weak.(2) The solution technique explained below applies to almost all weak bases. The only things to change are the concentration and the K

(b) The K_{b}value of a weak base is small (10¯^{5}is a typical value). Strong bases have very large K_{b}values (10^{5}or 10^{7}being typical values)._{b}, if doing another base.

Comment on 1b: This means K_{b} values of strong bases are billions and trillions of times larger than K_{b} values of weak bases. This means the reactions for strong bases essentially go completely to the right. In other words, strong bases (and acids) are 100% ionized in solution.

**Solution:**

Warning: long solution, some detailed explanation

1) From above, here is the K_{b} expression for ammonia:

[NH _{4}^{+}] [OH¯]K _{b}=––––––––––– [NH _{3}]

2) The key quantity we want is the [OH¯]. Once we have that, then the pH is easy to calculate. Since we do not know the value, let's do this:

[OH¯] = x

3) I hope that, right away, you can see this:

[NH_{4}^{+}] = x

This is because of the one-to-on molar ratio between OH¯ and NH_{4}^{+}, both being created as one NH_{3} molecule ionizes (see the coefficients of the balanced chemical equation above).

The end result of this is that [OH¯] will equal [NH_{4}^{+}] in the solution.

4) So now, we have all but one value in our equation:

(x) (x) 1.77 x 10¯ ^{5}=–––––– [NH _{3}]All we have to do is figure out [NH

_{3}] and we can calculate an answer to 'x.'

5) In this problem, the [NH_{3}] started at 0.100 M and went down as NH_{3} molecules reacted. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.100 − x. That mean this is the final set-up:

(x) (x) 1.77 x 10¯ ^{5}=–––––––– 0.100 − x That is a quadratic equation and can easily be solved with the quadratic formula. However, there is a trick we can use to make our calculation easier.

6) It turns out that K_{b} values are very difficult to figure out.There's a whole bunch of variables that are difficult to control. The end result is that K_{b} are approximate and most are in error about ± 5%.

So that means, if we stay within 5% of the answer using the quadratic, we can use approximate techniques to get an answer. The approximation occurs with '0.100 − x.' Since x is rather small, it will not change the value of 0.100 by much, so we can say:

0.100 − x ≈ 0.100

7) We now write a new equation:

(x) (x) 1.77 x 10¯ ^{5}=–––––––– 0.100

8) We move the 0.100 to the other side to get:

x^{2}= 1.77 x 10¯^{6}

9) Taking the square root (of both sides!!), we get:

x = 1.33 x 10¯^{3}M

10) Take note of two things:

(a) The K_{b}value is unitless, but x is a molarity. Some instructors insist on units for the K_{b}.

(b) Square root both sides. I have had students square root the x^{2}, but not the other side. Weird, but true.

11) We finish by determining the pOH first and then the pH:

pOH = −log [OH¯]pOH = −log 1.33 x 10¯

^{3}pOH = 2.876

pH + pOH = 14

pH + 2.876 = 14

pH = 11.124

We could have determined the [H

_{3}O^{+}] first (using the K_{w}rather than the pK_{w}), then determined the pH from the [H_{3}O^{+}].

12) The final step has to do with checking for 5%. The formula is:

([OH¯] / [NH_{3}]_{o}) (100) < 5%Where [NH

_{3}]_{o}is the starting concentration of the acid.In our case, we had 1.33%, which is acceptable.

There is a brief discussion of the 5% Rule here.

**Example #2:** What is the pH of a 0.300 M solution of morphine? K_{b} = 1.62 x 10¯^{6}

**Solution:**

1) Write the chemical reaction and the K_{b} expression:

M + H_{2}O ⇌ MH^{+}+ OH¯

M refers to the entire morphine molecule and MH

^{+}refers to the molecule after accepting a proton. It is completely unimportant what its formula is.

[MH ^{+}] [OH¯]K _{b}=––––––––––– [M]

2) We want the [OH¯]. Since it is an unknown value, we do this:

[OH¯] = x

3) This also is true:

[MH^{+}] = xThis is because of the one-to-one molar ratio between OH¯ and MH

^{+}, both being created as one M molecule ionizes (see coefficients of the balanced chemical equation above).The end result of this is that [OH¯] will equal [MH

^{+}] in the solution.

4) So now, we have all but one value in our equation:

(x) (x) 1.62 x 10¯ ^{6}=–––––– [M] All we have to do is figure out [HAc] and we can calculate an answer to 'x.'

5) In this problem, the [M] started at 0.300 M and went down as M molecules reacted with water. In fact, due to the one-to-one ratio, it went down by 'x' amount and wound up at an ending value of 0.300 − x. That mean this is the final set-up:

(x) (x) 1.62 x 10¯ ^{6}=–––––– 0.300 − x That is a quadratic equation and can easily be solved with the quadratic formula. However, there is a trick we can use to make our calculation easier.

6) It turns out that K_{b} values are very difficult to figure out.There's a whole bunch of variables that are difficult to control. The end result is that K_{b} are approximate and most are in error about ± 5%.

So that means, if we stay within 5% of the answer using the quadratic, we can use approximate techniques to get an answer. The approximation occurs with '0.100 − x.' Since x is rather small, it will not change the value of 0.100 by much, so we can say:

0.300 − x ≈ 0.300

7) We now write a new equation:

(x) (x) 1.62 x 10¯ ^{6}=–––––––– 0.300

8) We move the 0.300 to the other side to get:

x^{2}= 4.86 x 10¯^{7}

9) Taking the square root (of both sides!!), we get:

x = 6.97 x 10¯^{4}M

10) Take note of two things:

(a) The K_{b}value is unitless, but x is a molarity. Some instructors insist on units for the K_{b}.

(b) Square root both sides. I have had students square root the x^{2}, but not the other side. Weird, but true.

11) We finish by determining the pOH first and then the pH:

pOH = −log [OH¯]pOH = −log 6.97 x 10¯

^{4}pOH = 3.157

pH + pOH = 14

pH + 3.157 = 14

pH = 10.843

We could have determined the [H

_{3}O^{+}] first (using the K_{w}rather than the pK_{w}), then determined the pH from the [H_{3}O^{+}].

12) The final step has to do with checking for 5%. The formula is:

([OH¯] / [M]_{o}) (100) < 5%Where [M]

_{o}is the starting concentration of the acid.In our case, we had 0.23%, showing that the approximation was valid.

**Example #3:** What is the pH of a 0.250 M solution of strychnine? K_{b} = 1.82 x 10¯^{6}

1) You may have noticed that the solutions in Example #1 and #2 were exactly the same. Both ended up with this:

x = $\sqrt{\mathrm{(Kb)\; ([B]o)}}$where [B]

_{o}is the starting concentration of the base.When you're doing the 'drop subtract x' approximation, this equation just above always works for

MONOPROTICweak bases. Dibasic and tribasic bases have different solving techniques. Those techniques will not be covered.To carry the similarity one step farther, you may have noticed the similiar wordings in the K

_{a}examples and the K_{b}examples. I wrote the K_{a}examples and then just edited a copy using some K_{b}examples. Just remember, when doing a K_{b}problem, you wind up with the pOH and you have to do one more step involving pH + pOH = 14.

2) Solve the example:

x = $\sqrt{\mathrm{(1.82\; x\; 10\xaf6)\; (0.250)}}$x = 6.74 x 10¯

^{4}MpOH = 3.171

pH = 10.829

3) Checking the 5% rule, we get 0.16%. Using the approximate technique is valid.

**Example #4:** Determine the pH of a 2.54 M methylamine solution, CH_{3}NH_{2}. The K_{b} value of the weak base methyl amine is 5.0 x 10¯^{4}.

**Solution:**

1) Chemical equation and K_{b} expression:

CH_{3}NH_{2}+ H_{2}O ⇌ CH_{3}NH_{3}^{+}+ OH¯

[CH _{3}NH_{3}^{+}] [OH¯]K _{b}=––––––––––––––– [CH _{3}NH_{2}]

2) Solve for the hydroxide ion concentration via the shortcut used in Example #3:

x = $\sqrt{\mathrm{(Kb)\; ([B]o)}}$where [B]

_{o}is the starting concentration of the base.x = $\sqrt{\mathrm{(5.0\; x\; 10\xaf4)\; (2.54)}}$

x = 0.035637 M (this is the hydroxide concentration)

3) Determine the pOH, then the pH:

pOH = −log 0.035637 = 1.45pH = 14 − 1.45 = 12.55

4) 5% rule yields 1.4%. The approximate technique is valid.

**Example #5:** Codeine (C_{18}H_{21}NO_{3}) is a weak organic base. A 5.0 x 10¯^{3} M solution of codeine has a pH of 9.95. Calculate the value of K_{b} for this substance.

Comment: note that I will use B to symbolize the weak base. No one cares what the specific base is because the technique to be explained works for all weak bases. What happens is that some teachers will use the name of a specific weak base while others go the generic route.

**Solution:**

1) Write the ionizaton equation for the base. Remember, we will use B to symbolize the base.

B + H_{2}O ⇌ HB^{+}+ OH¯

2) Write the equilibrium expression:

K_{b}= ( [HB^{+}] [OH¯] ) / [B]

3) Our task now is to determine the three concentrations on the right-hand side of the equilibrium expression since the K_{b} is our unknown.

(a) We will use the pH to calculate the [OH¯]. We know pH = −log [H_{3}O^{+}], therefore [H_{3}O^{+}] = 10¯^{pH}[H

_{3}O^{+}] = 10¯^{9.95}= 1.122 x 10¯^{10}MI've kept a couple guard digits; I'll round off the final answer to the proper number of significant figures.

Knowing the [H

_{3}O^{+}] allows us to get the [OH¯]. To do this, we use K_{w}= [H_{3}O^{+}] [OH¯], so we have this:1.00 x 10¯

^{14}= (1.122 x 10¯^{10}) (x)x = [OH¯] = 8.9125 x 10¯

^{5}M(See note at bottom of this soluton for a different way to get the [OH¯].)

(b) From the ionization equation, we know there is a 1:1 molar ratio between [HB

^{+}] and [OH¯]. Therefore:[HB

^{+}] = 8.9125 x 10¯^{5}M

(c) the final value, [B] is given in the problem. In the example being discussed, 5.0 x 10¯^{3}M is the value we want. Some teachers will use 5.0 x 10¯^{3}M, while others would say to first subtract the 8.9125 x 10¯^{5}value from 5.0 x 10¯^{3}M. Let's do both.(i) K_{b}= [(8.9125 x 10¯^{5}) (8.9125 x 10¯^{5})] / 5.0 x 10¯^{3}K

_{b}= 1.59 x 10¯^{6}(II) K

_{b}= [(8.9125 x 10¯^{5}) (8.9125 x 10¯^{5})] / (5.0 x 10¯^{3}minus 8.9125 x 10¯^{5})K

_{b}= 1.62 x 10¯^{6}

In reality, it makes very little difference if we use the unmodified concentration of the acid or if we do the subtration. The percent difference between the two K_{b}values is 1.84%, well within the 5% rule to determine if the approximate technique is valid. And, in any event, both values round off to 1.6 x 10¯^{6}. In the end, you do what your teacher recommends. So, ask your teacher if you're not sure.

4) A different path to the [OH¯]:

pH = 9.95pH + pOH = pK

_{w}= 14.00pOH = 14.00 - 9.95 = 4.05

[OH¯] = 10¯

^{pOH}= 10¯^{4.05}= 8.9125 x 10¯^{5}M

In this example, I'll use a generic acid and made-up numbers that lead to the K_{b} for the base.

Before that, a comment: one reason teachers might tend to avoid real substances in this type of question is that you can just look up the answers on the Internet. K_{a} and K_{b} values for many weak acids and bases are widely available.

**Example #6:**

A 0.0135 M solution of a weak base (generic formula = B) has a pH of 8.39. Calculate the K_{b}for this weak base.

To remind you, here is the ionization equation:

B + H

_{2}O ⇌ HB^{+}+ OH¯

**Solution:**

[H

_{3}O^{+}] = 10¯^{pH}= 10¯^{8.39}= 4.0738 x 10¯^{9}M[OH¯] = K

_{w}/ [H_{3}O^{+}] = 1.00 x 10¯^{14}/ 4.0738 x 10¯^{9}[OH¯] = 2.4547 x 10¯

^{6}MRemember, due to the 1:1 molar ratio that [HB

^{+}] = [OH¯]I will use 0.0135 M for [B].

K

_{b}= [(2.4547 x 10¯^{6}) (2.4547 x 10¯^{6})] / 0.0135K

_{b}= 4.46 x 10¯^{10}

**Example #7:** A student prepares a 0.15 M solution of a monoprotic weak acid and determines the pH to be 11.68. What is the K_{b} of this weak base?

**Solution:**

pH = 11.68, so pOH = 14 − pH = 14.00 − 11.68 = 2.32[OH¯] = 10¯

^{pOH}= 10¯^{2.32}= 4.7863 x 10¯^{3}M[HB

^{+}] = [OH¯] = 4.7863 x 10¯^{3}M[B] = 0.15

K

_{b}= [(4.7863 x 10¯^{3}) (4.7863 x 10¯^{3})] / 0.15K

_{b}= 1.53 x 10¯^{4}

Comment: if you don't know the formula of the weak base, that's OK. Simply use B as the formula. It does not matter what the anion portion is, it only matters that the base is weak.

Also, with regard to bases, keep in mind that the proper generic chemical equation to use is:

B + H_{2}O ⇌ HB^{+}+ OH¯

As a reminder, here is the proper generic equation for weak acids:

HA + H_{2}O ⇌ H_{3}O^{+}+ A¯

**Example #8:** A 1.3 M solution of the weak base methylamine (CH_{3}NH_{2}) has a percent ionization of 0.72%. What is its K_{b}?

**Solution:**

1) Write the ionization equation for methylamine and its K_{b} expression:

CH_{3}NH_{2}+ H_{2}O ⇌ CH_{3}NH_{3}^{+}+ OH¯Use B in place of CH

_{3}NH_{2}:K

_{b}= ([HB^{+}] [OH¯]) / [B]

2) Use percent ionization to determine [OH¯]:

(1.3) (0.0072) = 0.00936 MThis is also [HB

^{+}].

3) Calculate K_{b}:

K_{b}= [(0.00936) (0.00936)] / 1.3K

_{b}= 6.7 x 10^{-5}

**Example #9:** The pH of a 0.150 M solution of a weak base is 10.98. Calculate the pH of a 0.040 M solution of the base.

**Solution:**

1) Calculate the [OH¯] of 0.150 M solution:

pH + pOH = 14.00pOH = 14.00 − 10.98 = 3.02

[OH¯] = 10¯

^{pOH}[OH¯] = 10¯

^{3.02}= 0.00095499 M

2) Determine the K_{b}:

B + H_{2}O ⇌ HB^{+}+ OH¯K

_{b}= ([HB^{+}] [OH¯]) / [B]K

_{b}= [(0.00095499) (0.00095499)] / 0.150K

_{b}= 6.08 x 10¯^{6}

3) Calculate [OH¯] of 0.040 M solution:

K_{b}= ([HB^{+}] [OH¯]) / [B]6.08 x 10¯

^{6}= [(x) (x)] / 0.040x = 0.000493 M

4) Calculate pOH, then pH of the 0.040 M solution

pOH = −log 0.000493 = 3.31pH = 14.00 − 3.31 = 10.69

**Example #10:** Cocaine is a weak organic base whose molecular formula is C_{17}H_{21}NO_{4}. An aqueous solution of cocaine was found to have a pH of 8.530 and an osmotic pressure of 52.7 torr at 15.0 °C. Calculate K_{b} for cocaine.

**Solution:**

1) The chemical reaction is this:

C_{17}H_{21}NO_{4}(aq) + H_{2}O(ℓ) ⇌ C_{17}H_{21}NO_{4}H^{+}(aq) + OH¯(aq)

2) Determine the concentration of cocaine:

PV = nRT

n P ––– = –––– V RT

0.069342 atm [C _{17}H_{21}NO_{4}]= ––––––––––––––––––––––––––– (0.08206 L atm / mol K) (288 K) [C

_{17}H_{21}NO_{4}] = 0.0029341 M

3) Determine the [OH¯]:

pH + pOH = pK_{w}8.530 + pOH = 14.000

pOH = 5.470

[OH¯] = 10¯

^{pOH}[OH¯] = 10¯

^{5.470}[OH¯] = 3.38844 x 10¯

^{6}M

4) Determine [C_{17}H_{21}NO_{4}H^{+}]:

From the 1:1 stoichiometry, we know this:[C

_{17}H_{21}NO_{4}H^{+}] = 3.38844 x 10¯^{6}M

5) We are now ready to determine the K_{b}:

[C _{17}H_{21}NO_{4}H^{+}] [OH¯]K _{b}=––––––––––––––––––– [C _{17}H_{21}NO_{4}]

(3.38844 x 10¯ ^{6}) (3.38844 x 10¯^{6})K _{b}=––––––––––––––––––––––––––– 0.0029341 K

_{b}= 3.91 x 10¯^{9}(to three sig figs)

**Bonus Example:** When 2.55 g of an unknown weak acid (MW: 85 g/mol) was added to 250. g of water, the freezing point of the resulting solution is −0.257 °C. What is the K_{b} of this acid?

**Solution:**

1) When the acid dissolves it ionizes. What we need to determine is the total amount of solute particles in the solution after it ionizes. We start with the freezing point depression equation:

ΔT = i K_{f}m

2) Calculate the molality:

2.55 g / 85.0 g/mol = 0.0300 mol0.0300 mol / 0.250 kg = 0.120 m

3) Calculate the van 't Hoff factor:

0.257 °C = (i) (1.86 °C / m) (0.120 m)i = 1.151434 <--- I'll keep some guard digits

4) A key assumption is that the density of the solution is 1.00 g/mL. This allows me to treat the 0.120 molal solution as being 0.120 molar.

5) The van 't Hoff factor is reflective of the total number of solute particles in solution, i.e. the total concentration:

(0.120 M) (1.151434) = 0.138172 M<--- rounded off a bit

6) Now, switch to the behavior of the acid. I will call it HA:

HA + H_{2}O ⇌ H_{3}O^{+}+ A¯K

_{a}= ([H_{3}O^{+}] [A¯]) / [HA]

7) We MUST know all the concentrations in order to get the K_{a}. We can do that using the total concentration of the solute particles:

[H^{+}] + [A¯] + [HA] = 0.138172 M(x) + (x) + (0.120 − x) = 0.138172

x = 0.018172 M

8) We can now solve for the K_{a}:

K_{a}= [(0.018172) (0.018172)] / 0.101828K

_{a}= 0.003242935

9) We determine the K_{b} from this relationship:

K_{a}* K_{b}= K_{w}(0.003242935) (K

_{b}) = 1.00 x 10¯^{14}K

_{b}= 3.08 x 10¯^{12}

Go to a listing of many types of acid base problems and their solutions