Question

4.13

Just as pH is the negative logarithm of [H3O+], pKa is the negative logarithm of Ka,

pKa=−logKa

The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions:

pH=pKa+log[base][acid]

Notice that the pH of a buffer has a value close to the pKa of the acid, differing only by the logarithm of the concentration ratio [base]/[acid].

Part B

How many grams of dry NH4Cl need to be added to 2.10 L of a 0.400 M solution of ammonia, NH3, to prepare a buffer solution that has a pH of 8.68? Kb for ammonia is 1.8×10−5.

Express your answer numerically in grams to three significant figures.

4.15

The hydroxide ion has the formula OH−. The solubility-product constants for three generic hydroxides are given here.

Generic hydroxide	Ksp
XOH	2.10×10−8
Y(OH)2	1.60×10−10
Z(OH)3	5.50×10−15

Use these values to answer the following questions:

The removal of an ion is sometimes considered to be complete when its concentration drops to 1.00×10−6  M. What concentration of hydroxide would cause Y2+ to "completely" precipitate from a solution?

Answer 1

Solution :-

Q4.13 part B) Calculating the mass of ammonium chloride

pH= 8.68

kb = 1.8*10^-5

pkb= -log kb

pkb= - log [1.8*10^-5]

pkb= 4.74

pkb+ pka= 14

pka= 14 – pkb

pka = 14 – 4.74

pka= 9.26

2.10 L of 0.400 M NH3 solution

Using the Henderson equation we can find the concentration of the NH4^+ needed to get the desired product

pH= pka + log [base]/[acid]

pH= pka + log [NH3]/[NH4^+]

8.68 = 9.26 + log[0.400]/[NH4^+]

8.68-9.26 = log[0.400]/[NH4^+]

-0.58 = log[0.400]/[NH4^+]

10^(-0.58) = [0.400]/[NH4^+]

0.263 = [0.400]/[NH4^+]

[NH4^+] = 0.400 / 0.263

[NH4^+] = 1.52 M

So now using the molarity of the ammonium ion and volume of solution we can find the moles of NH4Cl needed

Moles= molarity x volume in liter

Moles of NH4^+ = 1.52 mol per L * 2.10 L = 3.192 mol NH4^+

1 mol NH4^+ = 1 mol NH4Cl

Therefore moles of NH4Cl are same as moles of NH4^+

Hence moles of NH4Cl = 3.192 mol

Mass = moles x molar mass

Mass of NH4Cl = 3.192 mol * 53.491 g/mol

                           = 171 g NH4Cl

Therefore we need to add 171 g ammonium chloride (NH4Cl)

Q4.15)

Y(OH)2 ---- > Y^2+ + 2OH^-                     Ksp= 1.60*10^-10

Lets calculate the concentration of the OH^- ion when Y^2+ is 1*10^-6 M

Ksp = [Y^2+][OH^-]^2

Ksp =[x][2x]^2

1.60*10^-10 = [1.00*10^-6] [2x]^2

1.60*10^-10 = [1.00*10^-6] 4x^2

1.60*10^-10 / 1.00*10^-6 = 4x^2

1.60*10^-4 = 4x^2

1.60*10^-4 / 4 = x^2

4.00*10^-5 = x^2

(4.00*10^-4)^(1/2) = x

6.33*10^-3 = x

Therefore the concentration of the [OH^-] needed is 2x

[OH^-] = 2x

            = 2*6.33*10^-3 M

            = 1.26*10^-2 M

So we need hydroxide ion concentration 1.26*10^-2 M to precipitate out all the Y^2+