# Question #0bc73

##### 2 Answers

#### Explanation:

**!! EXTREMELY LONG ANSWER !!**

You're dealing with a **neutralization reaction** in which **weak base**, reacts with hydrochloric acid, **strong acid**, to produce **conjugate acid**, and water.

Your strategy here will be to write the balanced chemical equation for the two reactions between the weak base and the strong acid, and use equation stoichiometry to try and find *two relationships* between the two expressions of the *base dissociation constant*,

Since hydrochloric acid is a strong acid, it will dissociate completely in aqueous solution to form hydronium cations, **mole ratio**.

The chloride anions are *spectator ions*, so I will exclude them from the balanced chemical equations.

Now, let's assume that the **moles** of

Use the *molarity* of the hydrochloric acid solution, which will be **equivalent to its given normality**, to calculate how many *moles* of strong acid are present in the

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_(H_3O^(+)) = "0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(5.00 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))#

# = 5.0 * 10^(-4)"moles H"_3"O"^(+)#

The balanced chemical equation for this reaction looks like this

#" " "BOH"_ ((aq)) + "H"_ 3"O"_ ((aq))^(+) -> "B"_ ((aq))^(+) + 2"H"_ 2"O"_((l))#

The weak base reacts with the hydronium cations produced by the hydrochloric acid in a **mole ratio**. This means that **every mole** of weak base that takes part in the reaction consumes **one mole** of strong acid and produces **one mole** of conjugate acid.

You know for a fact that the strong acid will be **completely consumed** by the reaction, since the pH of the solution after the addition of the

So, if

#n_(H_3O^(+)) = "0 moles H"_3"O"^(+)-># completely consumed

#n_(BOH) = (x - 5.0 * 10^(-4))color(white)(a)"moles BOH"#

#n_(B^(+)) = 0 + 5.0 * 10^(-4) = 5.0 * 10^(-4)color(white)(a)"moles B"^(+)#

The **total volume** of the solution will be

#V_"total" = "40 mL" + "5.0 mL" = "45 mL"#

The **concentrations** of the weak base and conjugate acid will thus be

#["BOH"] = (x - 5.0 * 10^(-4)"moles")/(45 * 10^(-3)"L") = (x-5.0 * 10^(-4)) * 10^3/45color(white)(a)"mol L"^(-1)#

#["B"^(+)] = (5.0 * 10^(-4)"moles")/(45 * 10^(-3)"L") = 1.11 * 10^(-2)color(white)(a)"mol L"^(-1)#

Now, the dissociation of the weak base can be described by the following *equilibrium reaction*

#"BOH"_ ((aq)) rightleftharpoons "B"_ ((aq))^(+) + "OH"_((aq))^(-)#

Use the pH of the solution to find its pOH

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#

In this case, you will have

#"pOH" = 14 - 10.04 = 3.96#

As you know, the concentration of hydroxide anions can be found using

#color(purple)(|bar(ul(color(white)(a/a)color(black)(["OH"^(-)] = 10^(-"pOH"))color(white)(a/a)|)))#

This will get you

#["OH"^(-)] = 10^(-3.96) = 1.10 * 10^(-4)"M"#

By definition, the *base dissociation constant* will be equal to

#K_b = (["B"^(+)] * ["OH"^(-)])/(["BOH"])#

Plug in these values to get

#K_b = (1.11 * 10^(-2) * 1.10 * 10^(-4))/((x - 5.0 * 10^(-4)) * 10^3/45) = (5.5 * 10^(-8))/(x - 5.0 * 10^(-4))" " " "color(orange)((1))#

Now do the same thing for the second hydrochloric acid sample. Keep in mind that you're **not adding** a *separate sample* of

The **total volume** of acid added to the weak base solution will be equal to **adding**

This sample will contain

#n_(H_3O^(+)) = "0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))#

# = 1.5 * 10^(-3)"moles H"_3"O"^(+)#

Once again, the moles of hydronium ions coming from the acid will be **completely consumed** by the reaction. This time, the solution already contained **moles** of **moles** of

#n_(H_3O^(+)) = "0 moles H"_3"O"^(+)-># completely consumed

#n_(BOH) = x - 5.0 * 10^(-4) - 1.5 * 10^(-3) = (x - 2.0 * 10^(-3))color(white)(a)"moles BOH"#

#n_(B^(+)) = 5.0 * 10^(-4) + 1.5 * 10^(-3) = 2.0 * 10^(-3)color(white)(a)"moles B"^(+)#

The **total volume** of the solution will now be

#V_"total" = "45 mL" + "15 mL" = "60 mL"#

The pOH of the solution, knowing that the pH of the solution is equal to

#"pOH" = 14 - 9.14 = 4.86#

This means that you have

#["OH"^(-)] = 10^(-4.86) = 1.4 * 10^(-5)"M"#

The concentrations of the weak base and of its conjugate acid will now be

#["BOH"] = (x - 2.0 * 10^(-3)"moles")/(60 * 10^(-3)"L") = (x-2.0 * 10^(-3)) * 10^3/60color(white)(a)"mol L"^(-1)#

#["B"^(+)] = (2.0 * 10^(-3)"moles")/(60 * 10^(-3)"L") = 3.33 * 10^(-2)color(white)(a)"mol L"^(-1)#

The base dissociation constant will be equal to

#K_b = (3.33 * 10^(-2) * 1.4 * 10^(-5))/((x - 2.0 * 10^(-3)) * 10^3/60) = (2.8 * 10^(-8))/(x - 2.0 * 10^(-3))" " " "color(orange)((2))#

You now have two equations with two unknowns,

#color(orange)((1)) = color(orange)((2)) implies (5.5 * color(red)(cancel(color(black)(10^(-8)))))/(x - 5.0 * 10^(-4)) = (2.8 * color(red)(cancel(color(black)(10^(-8)))))/(x - 2.0 * 10^(-3))#

Rearrange to get

#5.5 * (x - 2.0 * 10^(-3)) = 2.8 * (x - 5.0 * 10^(-4))#

#5.5x - 2.8x = 11 * 10^(-3) - 1.4 * 10^(-3)#

#x = (9.6 * 10^(-3))/2.7 = 3.56 * 10^(-3)#

Plug this into equation

#K_b = (5.5 * 10^(-8))/(3.56 * 10^(-3) - 5.0 * 10^(-4)) = color(green)(|bar(ul(color(white)(a/a)1.8 * 10^(-5)color(white)(a/a)|)))#

I'll leave the answer rounded to two **sig figs**.

This is an alternative to the answer originally provided by Stefan V. This solution utilizes the properties of buffer solutions.

#### Explanation:

We know that because the pH remains above 7.0, the neutralization is only partial, and therefore we have a buffer solution. For any buffer consisting of a conjugate acid-base pair, we know that

or

Here, we'll use the second version, where

Let

After the first addition of

we have

and

Similarly, after the second addition of

and

We now have 2 equations and two unknowns, so subtracting the first from the second gives

Take the anti-log of both sides and rearrange to give

Plugging this back into either of the pOH equations above gives

or