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?
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+
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