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Calculate K_{a} from pH and Molarity

This is a favorite problem for teachers to test! Here's the problem to be discussed:

**Problem #1:** 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^{+}], therefore [H^{+}] = 10¯^{pH}[H

^{+}] = 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

^{+}] allows us to get the [OH¯]. To do this, we use K_{w}= [H^{+}] [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.c1) K_{b}= [(8.9125 x 10¯^{5}) (8.9125 x 10¯^{5})] / 5.0 x 10¯^{3}K

_{b}= 1.59 x 10¯^{6}c2) 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 (the 5.0 x 10¯^{3} value) or if we do the subtration. In the above example, 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.

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.

**Problem #2:**

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:**

a) [H

^{+}] = 10¯^{pH}= 10¯^{8.39}= 4.0738 x 10¯^{9}Mb) [OH¯] = K

_{w}/ [H^{+}] = 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¯]c) I will use 0.0135 M for [B].

K

_{b}= [(2.4547 x 10¯^{6}) (2.4547 x 10¯^{6})] / 0.0135 = 4.46 x 10¯^{10}

**Problem #3:** 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? The answer. In the answer, I plan to follow a slightly different path to the hydroxide ion concentration.

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^{+}+ A¯

**Problem #4:** A 0.15 mol/L solution of the ascorbate ion, HC_{6}H_{6}O_{6}¯, has a pH of 8.65. Calculate the K_{b} for the ascorbate ion.

**Solution:**

1) Ascorbate ion reacts as follows:

HC_{6}H_{6}O_{6}¯ + H_{2}O <===> H_{2}C_{6}H_{6}O_{6}+ OH¯

2) The K_{b} expression is:

K_{b}= ([H_{2}C_{6}H_{6}O_{6}] [OH¯]) / [HC_{6}H_{6}O_{6}¯]

3) Substituting values and solving:

K_{b}= [(4.4668 x 10¯^{6}) (4.4668 x 10¯^{6})] / 0.15K

_{b}= 1.33 x 10¯^{10}

Comment: use the pH to calculate the [H^{+}]. Then use K_{w} and the [H^{+}] to calculate the hydoxide ion concentration.

Calculate K_{a} from pH and Molarity

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