Answer to Question #171967 in General Chemistry for Irish

Question

Prepare 750 ml of a 1:1500 acetic acid solution. SG of acetic acid is 1.05; pka = 4.76. 

a. How many grams of acetic acid should be used?

b. what is the molality

c. what is the pH of the solution

Answer 1

Q171967

Prepare 750 ml of a 1:1500 acetic acid solution. SG of acetic acid is 1.05; pka = 4.76. 

a. How many grams of acetic acid should be used?

b. what is the molality

c. what is the pH of the solution.

Solution:

The volume ratio of acetic acid in the solution is 1: 1500.

This means there is 1mL of acetic acid in 1500 mL of solution. Using this ratio we can find the volume of acetic acid required for the preparation of 750 mL of solution.

Step 1: To find the volume of acetic acid required to prepare a 750 mL solution.

$$volume\space of\space acetic\space acid\space required = 750 mL\space solution * \frac{1\space mL\space acetic \space acid }{1500\space mL\space solution }$$

$= 0.50\space mL\space acetic\space acid.$

Step 2: To convert 0.50 mL of acetic acid to grams by using the given specific gravity.

the specific gravity of acetic acid = 1.05.

specific gravity is always calculated considering the water as a reference.

If we assume that the density of water is 1.00 g/mL.

Then the density of acetic acid = specific gravity * density of water

= 1.05 * 1.00 g/ mL

= 1.05 g/mL.

Now we will plug the density and volume of acetic acid in the density formula and find the

mass of acetic acid.

$$density = \frac{mass }{volume }$$

arranging the formula for mass we have

$$mass = density * volume$$

$mass\space of\space acetic\space acid =\space 1.05\space g/mL * 0.50\space mL = 0.525\space grams.$

Answer a: We will require 0.525 grams of acetic acid for preparing 750 mL of 1 : 1500 acetic acid solution.

b) What is the molality?

Answer :

The formula of molality is

$$Molality = \frac{moles\space of\space solute }{mass\space of\space solvent\space in\space 'kg' }$$

acetic acid is the solute and water is the solvent.

volume of solution = 750 mL.

volume of acetic acid = 0.50 mL.

volume of solution = volume of acetic acid + volume of water.

750 mL = 0.50 mL + volume of water.

So volume of water = 750 mL - 0.50 mL = 749.5 mL .

density of water is 1.00 g/mL.

so mass of water = density * volume = 1.00 g/mL * 749.5 mL

= 749.5 grams.

Convert this to kilograms by using the conversion factor, 1kg = 1000 grams.

$$mass\space of\space water\space in\space 'kg' = 749.5\space g * \frac{1\space kg }{1000\space g }$$

$$= 0.7495 kilograms.$$

We will also require moles of acetic acid.

Convert 0.525 grams of acetic acid to moles by using the molar mass of acetic acid.

molar mass of CH3COOH = 2 * atomic mass of C + 2 * atomic mass of O + 4 * atomic mass of H

= 2 * 12.011 g/mol + 2 * 15.999 g/mol + 4 * 1.00794 g/mol

= 24.022 g/mol + 31.998 g/mol + 4.03176 g/mol

= 60.052 g/mol

$$moles\space of\space acetic\space acid = 0.525 g\space CH_3COOOH * \frac{1\space mol\space CH_3COOOH }{60.052\space g\space of\space CH_3COOOH}$$

$= 0.008742\space mole\space of\space CH_3COOH.$

substitute, moles of acetic acid = 0.008742 mol and mass of water = 0.7495 kg in the

morality formula we have

$$Molality = \frac{0.008742\space mol\space CH_3COOH }{0.7495\space kg }$$

$$= 0.01166 molal.$$

which in 3 significant figures is 0.0117 molal.

Hence the molality of the acetic acid solution is 0.0117 molal.

c) What is the pH of the solution?

Solution :

The dissociation of acetic acid is given as

CH₃COOH (aq) <==> CH₃COO^- (aq) + H⁺ (aq)

The molarity of the acetic acid solution would too be 0.0117 M.

This is because the given solution is very dilute.

Using this draw the ICE table.

$$CH_3COOH (aq) <==> CH_3COO- (aq) + H+ (aq)$$

Initial 0.0117 M 0 0

Change - x + x + x

Equilibrium 0.0117 - x x x

The Keq equation for the given reaction is written as

$$Ka = \frac{[CH_3COO^-] [H^+] }{[CH_3COOH] };$$

We are given, pKa = 4.76.

pKa = - log [Ka]

4.76 = -log [Ka]

Ka = 10^-4.76 = 1. 738 * 10^-5 ;

Substitute the equilibrium concentration and Ka in the formula we have

$$1.738 * 10^{-5} = \frac{x * x }{0.0117 - x };$$

here 'x' is very small compared to the concentration of acetic acid.

so 0.0017 - x is approximately equal to 0.00117M

x² = 1.738 * 10^-5 * 0.0117 = 2.033 * 10^-7

Taking square root we have

x = 4.4423 * 10^-4 M .

[H+] = x = 4.4423 * 10^-4 M .

pH = -log[ H+] = -log [ 4.4423 * 10^-4 M ] = 3.35 ;

Hence the pH of the solution would be 3.35.