Selective Preciptation Selective Precipitation Part A Consider the reaction Erte

Question

Selective Preciptation Selective Precipitation Part A Consider the reaction Erter the K expression for AaB,() in terms of IAI and IB Express your answer in terms of IA and BI, for example, IAJ 2 The reaction quotient. Q, saso eouNtoIVIIZI Q only equilibrum. When Q is equal than the soution is unsaturated when Qis gverter than K the solution is supersaturted Md Hot MYAtsmen Give up Review Part precipitation oould ooour Incorrect: Try Again Part B Emer the K. .pression for AC. ntems of IA and Express your answer in terms ofIAl and ICI, for example, (x] [Yl'z Hints Manswers Give up Review Pat Incorrect: Try Again EE OAsk me anything

Explanation / Answer

PART A:

Henderson-Hesselbalach equation is expressed as

pH = pKa + log { [salt] / [acid] }

pKa of HClO = 7.54

[salt] = [KClO] = 0.155 M

[acid] = [HClO] =0.115 M

So, pH = 7.54 + log (0.155 / 0.115)

= 7.54 + log (1.35)

= 7.54 + 0.13

= 7.67

PART B:

This is a buffer solution since it contains a weak acid (C2H5NH3+ in the form of C2H5NH3Br) and its conjugate base, C2H5NH2. The pH is readily found using the Henderson-Hasselbalch equation:

pH = pKa + log (moles buffer base / moles buffer acid)

Since the Kb for C2H5NH2 = 5.6 x 10-4, then

Ka C2H5NH3+ = Kw / Kb = (1 x 10-14) / (5.6 x 10-4) = 1.8 x 10-11

pKa = -log Ka = - log (1.8 x 10-11) = 10.74

pH = 10.74 + log (moles C2H5NH2 / moles C2H5NH3Br)

All we need is the mole ratio of C2H5NH2 to C2H5NH3Br.

Let's say we have 100 g of solution. Then based on the percentages given, there are 1.45 g of C2H5NH2 and 1.15 g of C2H5NH3Br.

1.45 g C2H5NH2 x (1 mole C2H5NH2 / 45.1 g C2H5NH2) = 0.0322 moles C2H5NH2

1.15 g C2H5NH3Br x (1 mole C2H5NH3Br / 126.0 g C2H5NH3Br) = 0.0091 moles C2H5NH3Br

pH = 10.74 + log (0.0322 / 0.0091)

      = 10.74 + 0.55

      = 11.29

PART C:

This is a buffer solution since it contains a weak acid CH3COOH and its conjugate base CH3COONa. The pH is readily found using the Henderson-Hasselbalch equation:

Molar mass of CH3COOH = 60 g/mol

So, 13.5 g of CH3COOH = (13.5 / 60) mol = 0.225 mol

Molar mass of CH3COONa = 82 g/mol

So, 15 g of CH3COONa = (15 / 82) mol = 0.183 mol

Volume = 150 mL = 0.150 L

So

[CH3COOH] = 0.225 mol / 0.150 L = 1.50 M
[CH3COONa] = 0.183 mol / 0.150 L = 1.22 M

pH = pKa + log ([salt] / [acid] )

Ka for CH3COOH = 1.8 x 10-5, then

pKa = -log Ka = - log (1.8 x 10-5) = 4.76

pH = pKa + log ([CH3COONa] / [CH3COOH] )

= 4.76 + log (1.22 / 1.50)

      = 4.76 – 0.09

      = 4.67

PART D:

0.120 L x 0.105 M NH3 = 0.0126 mol of NH3
0.120 L x 0.135 M NH4Br = 0.0162 mol of NH4Br

Kb NH3 = 1.8 x 10-5

Ka = (1.0 x 10-14 ) / (1.8 x 10-5 ) = 5.56 x 10-10

pKa = - log Ka = - log(5.56 x 10-10) = 9.25

pH = pka + log [ NH3 / NH4+]

When HCl is added, it will react with NH3 to form NH4+. So, moles of NH3 will decrease and moles of NH4+ will increase.

Now, for pH = 9, we can write

pH = pka + log [ NH3 / NH4+]

9.00 = 9.25 + log { (0.0126 - x) / (0.0162 + x) }

- 0.25 = log { (0.0126 - x) / (0.0162 + x) }

(0.0126 - x) / (0.0162 + x) = 10-0.25

(0.0126 - x) / (0.0162 + x) = 0.56

(0.0126 - x) = 0.56 (0.0162 + x)

0.0126 - x = 0.0091 + 0.56x

1.56x = 0.0035

x = 0.0022 mole

Molar mass of HCl = 36.46 g/mole

So, 0.0022 mole of HCl = 0.0022 x 36.46 g = 0.08 g

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