What is pH of a 400 ml aqueous solution that contains 1.0 g potassium hydroxide (KOH).

Number of moles = mass / molar mass

Molar mass (molecular weight) of KOH = 56.11 g/ mole

Number of moles = 1.0 g/ (56.11 g/ mole) = 0.018 mole

Molar concentration = number of moles/ volume of solution in liters

Volume of solution in liters = 400 ml/ 1000 = 0.4 L

Molar concentration = 0.018 mole/ (0.4 L) = 0.045 M

pOH = -log [OH^-]

[OH^-] = 0.045 M

pOH = - log 0.045 = 1.35

pH + pOH =14

pH = 14 - pOH = 14 - 1.35 = 12.65

Molar concentration (molarity) of 1% (w/v) borax solution.

1% (w/v) = 1 g/ 100 ml = 10 g/ 1000 ml = 10 g/ 1 liter

Molar mass (molecular weight) of borax = 381.37 g/ mole.

                                 Molar concentration = number of moles/ volume of solution in liters

Number of moles = mass/ molar mass

                             = 10 g/ (381.37 g/ mole)

                             = 0.026 mole

Molar concentration = 0.026 mole/ 1 liter = 0.026 M.

 

What is the ratio of the molar concentrations of acetic acid to sodium acetate ([acetic acid]/ [sodium acetate]) required to prepare a buffer solution with a pH value equal to 4.5.

Solution:

pH = pKa + log ([sodium acetate]/ [acetic acid])

pKa = -log Ka

Assuming that the acid dissociation (Ka) value for acetic acid is 1.75 × 10^-5 at 25 ºC.

pKa = - log (1.75 × 10^-5) = 4.76

4.5 = 4.76 + log ([sodium acetate]/ [acetic acid])

log ([sodium acetate]/ [acetic acid]) = 4.5 - 4.76 = -0.26

[sodium acetate]/ [acetic acid] = 10^-0.26 = 0.55

[acetic acid]/ [sodium acetate] = 1/ 0.55 = 1.82

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What is buffer capacity?

What is molar concentration (molarity) of 20% (weight/ volume) glycerin solution.

Solution:

Molar concentration = number of moles/ volume of solution in liters.

20% = 20 grams/ 100 ml solution.

20% = 200 grams/ 1000 ml solution = 200 g/ liter.

Number of moles = mass/ molar mass

Molar mass ( molecular weight ) of glycerin (glycerol) = 92.1 g/ mole.

Number of moles = 200 g/ (92.1 g/ mole) = 2.17 mole.

Molar concentration = 2.17 mole/ 1 liter = 2.17 M.

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Calculate molar concentration (molarity) of 0.9% (weight/ volume) sodium chloride (NaCl) solution.

Solution:

Molar concentration = number of moles/ volume of solution in liters

0.9% = 0.9 grams/ 100 ml solution.

0.9% = 9 grams/ 1000 ml solution = 9 g/ liter.

Number of moles = mass/ molar mass

Molar mass ( molecular weight ) of sodium chloride = 58.44 g/ mole.

Number of moles = 9 g/ (58.44 g/ mole) = 0.15 mole.

Molar concentration = 0.15 mole/ 1 liter = 0.15 M.

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Calculate pH of a 500 ml solution containing 0.6 grams sodium hydroxide (NaOH).

Solution:

Step 1. Calculate molar concentration.

Molar concentration = number of moles/ volume in liters

Number of moles = mass/ molar mass

Molar mass (molecular weight) of sodium hydroxide = 40.0 g/ mole.
Volume of solution in liters = 500 ml/ (1000 ml/ liter) = 0.5 liter.

Number of moles = 0.6 g / (40.0 g/ mole) = 0.015 mole.

Molar concentration = 0.015 mole/ (0.5 liter) = 0.03 M.

Step 2. Calculate pH using molar concentration.

pOH = - log [OH^-]

pOH = - log 0.03 = 1.52

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 1.52 = 12.48

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What is pH of a 1 liter solution containing 0.5 grams of potassium hydroxide (KOH)?

Solution:

Step 1. Calculate molar concentration.

Molar concentration = number of moles/ volume in liters

Number of moles = mass/ molar mass

Molar mass (molecular weight) of potassium hydroxide = 56.1 g/ mole.

Number of moles = 0.5 g / (56.1 g/ mole) = 0.00891 mole.

Molar concentration = 0.00891 mole/ (1 liter) = 0.00891 M.

Step 2. Calculate pH using molar concentration.

pOH = - log [OH^-]

pOH = - log 0.00891 = 2.05

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 2.05 = 11.95

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What is pH of 0.03 M hydrochloric acid (HCl) solution.

Hydrochloric acid is a strong acid that dissociates in water to produce chloride and hydronium (H₃O⁺) ions. The pH of a strong acid solution can be calculated using the following equations:

pH = - log [H₃O⁺]

[H₃O⁺] = [strong acid] = 0.03

pH = - log 0.03 =1.52

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What is pH of 0.05 M potassium hydroxide solution (KOH) solution?

Potassium hydroxide is a strong base that ionizes in water to produce potassium and hydroxyl (OH^-) ions. The following equations can be used to calculate pH for a strong base solution:

pOH = - log [OH^-]

[OH^-] = [strong base] = 0.05 M

pOH = - log 0.05 = 1.30

pH + pOH =14.00

pH = 14.00 - pOH

pH = 14.00 - 1.30 = 12.70

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Prepare 1 liter acetic acid solution with pH=5.1.

Acetic acid is a weakly acidic compound that dissociates in solution to produce sodium acetate and hydronium ions ([H₃O⁺]). The following equations can be used to estimate pH of acetic acid solution:

 

[H₃O⁺] = [H⁺]

Where, Ka is acid dissociation constant of the acid, and C is the molar concentration of the acid.

Solution:

Step 1. apply the above equations to find molar concentration: 

pH = -log [H₃O⁺]

5.1 = -log [H₃O⁺]

[H₃O⁺] = 10^-5.1

[H₃O⁺] = 7.94 × 10^-6

Assuming Ka value for acetic acid at 25 ºC is equal to 1.75 × 10^-5 then the above equation can be written as the following:

7.94 × 10^-6 =((-1.75 × 10^-5 + sqrt ((1.75 × 10^-5)2+ 4 ×1.75 × 10^-5× C))/2

  

Solving the above equation for C gives the following value:

C= 1.15 × 10^-5 M

Step 2. Find out if preparation and dilution of a stock solution is required.

Molecular weight (molar mass) of acetic acid is equal to 60.1 g/ mole. 

To prepare 1 liter of 1.15 × 10^-5 M we need 1.15 × 10^-5 mole of acetic acid. Therefore, the mass of acetic acid needed is calculated as the following:

Mass = number of moles × molar mass

Mass = 1.15 × 10^-5 × 60.1 = 0.00069 g = 0.69 mg.

The above mass is too small to be simply weighed on a scale. Therefore, a stock solution should be prepared then a certain volume of the stock solution containing the required number of moles is used in the preparation.

Step 3. Make calculations for stock solution and required dilution(s).

To prepare 0.115 M stock solution we need to dissolve 6.9 g acetic acid in water to make 1 liter solution. Calculate the volume of this stock solution which contains 

1.15 × 10^-5 mole using the following equation:

Volume = number of moles/ concentration

Volume = 1.15 × 10^-5 mole/ 0.115 M

Volume = 0.0001 liter

Volume = 0.0001 liter * 1000 ml/ liter = 0.1 ml.

The calculated volume of 0.1 ml is too small to be accurately taken. Therefore, further dilution is required as the following:

Take 10 ml (0.01 liter) from the stock solution to prepare 1 liter solution. The concentration of this solution is calculated as the following:

concentration = number of moles/ volume

concentration = (0.01 liter × 0.115 mole/ liter) / 1 liter

concentration = 0.00115 M

The diluted solution with concentration of 0.00115 M can be used to prepare the needed solution as the following:

volume = number of moles/ concentration

volume = 1.15 × 10^-5 mole/ 0.00115 M

volume = 0.01 liter

volume = 0.01 liter * 1000 ml/ liter = 10 ml.

Thus, add 10 ml of the diluted solution (0.00115 M) to a 1 liter volumetric flask then complete volume to 1 liter mark wit water and mix.

Examples on calculations for pharmaceutical isotonic solutions.

Example 1. Prepare 100 ml isotonic solution of 0.5 % (w/v) ephedrine sulfate (E value = 0.23).

Answer 1. Mass of ephedrine sulfate = concentration × volume = 0.5  (g/ 100 ml) × 100 ml = 0.5 g.
According to the E value each gram of ephedrine sulfate is equivalent to 0.23 g sodium chloride. Therefore, 0.5 g is equivalent to 0.23 × 0.5 = 0.115 g sodium chloride. The amount of sodium chloride needed to make the solution isotonic is equal to 0.9 - 0.115 = 0.785 g. The formula is written as the following:

Ephedrine sulfate        0.5 g
Sodium chloride         0.785 g
Water q.s.                 100 ml


Example 2. What mass of sodium chloride is included in 100 ml of 5.0 % (w/v) dextrose monohydrate (E value = 0.16) isotonic solution.

Answer 2. Mass of dextrose monohyrate in 100 ml = concentration × volume = 5.0 (g/ 100 ml) × 100 ml = 5.0 g. Each gram of dextrose monohydrate is equivalent to 0.16 g sodium chloride (E value = 0.16). Therefore, 5 g dextrose monohydrate is equivalent to 0.8 g (5 × 0.16) sodium chloride. To make 100 ml of this solution isotonic some amount of sodium chloride needs to be included such that total solute effect is equivalent to that of 0.9 g sodium chloride. Thus, the formula for 100 ml is written as the following:

Dextrose monohydrate 5.0 g
Sodium chloride           0.1 g (0.9 - 0.8)
Water q.s.                   100 ml

Example 3. Prepare an isotonic solution of ethanol (E value = 0.7).

Answer 3. Each gram of ethanol (anhydride) is equivalent to 0.7 g of sodium chloride. Thus, the amount needed for 100 ml solution in order to obtain an equivalent effect to that of 0.9 % sodium chloride is 1.29 g (0.9/ 0.7). Therefore, 1.29 % ethanol solution is isotonic.

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Calculate pH of 0.1 % (w/v) atropine solution.

General information:
Atropine (B) is a weakly basic compound. It partially ionizes in water  to produce protinated atropine (BH⁺)  and hydroxyl ions (OH^-) according to the following formula:

B + H₂O ↔ BH⁺ + OH^-

Assuming that it has Kb value of 4.5 × 10^-5 and a molar mass (molecular weight) equal to 289.4 g/ mole then pH of the solution can be calculated as the following.

Calculations:
Step 1. Calculation of molar concentration.
Molar concentration = (0.1g/ 100ml) × (1000 ml/ liter) × ( 1/ (289.4 g/ mole) = 0.0035 mole/ liter.

Step 2. Calculation of pH by applying the following equations:

[OH^-] = (-(4.5 × 10^-5) + sqrt((4.5 × 10^-5)²+(4 × 4.5 × 10^-5 × 0.0035)))/ 2
[OH^-] = 3.75 × 10^-4
pH = 14 + log (3.75 × 10^-4) = 10.57

Check that the pH of 0.0035 M atropine (Kb = 4.5 × 10^-5) solution is equal to 10.57 using online pH calculator.

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Calculate pH of a solution prepared by dissolving 0.09 g potassium hydroxide (KOH) to make 500 ml assuming that 7% of its mass is moisture.

General Information:
Potassium hydroxide (KOH) is a strong base. It completely ionizes in solution to produce potassium ions (K⁺) and hydroxyl ions (OH^-). It has a molar mass (molecular weight) equal to 56.11 g/ mole.

Calculations:

Mass of moisture = 0.09 × (7/100) = 0.0063 g
Actual KOH mass = 0.09 - 0.0063 = 0.0837 g
Number of moles = mass/ molar mass = 0.0837 ÷ (56.11 g/ mole) = 0.0015 mole
Volume of solution in liters = 500 ml × (0.001 liter/ ml) = 0.5 liter
Molar concentration = number of moles/ volume = 0.0015 mole / 0.5 liter = 0.003 moles/ liter

[OH^-] = [KOH]
pH = 14 + log (0.003)
pH = 14 - 2.52
pH = 11.48

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Calculate mass of potassium hydroxide (KOH) contained in 500 ml solution with pH = 11.5.

General Information:
Potassium hydroxide (KOH) is a strong base. It completely ionizes in solution to produce potassium ions (K⁺) and hydroxyl ions (OH^-). It has a molar mass (molecular weight) equal to 56.11 g/ mole.

Calculations:
pH = 14 + log [OH^-]
11.5 = 14 + log [OH^-]
-2.5 = log [OH^-]
[OH^-] = 10^-2.5 = 0.0032 mole/ liter.

[KOH] = [OH^-] = 0.0032 mole/ liter.

Volume of solution in liters = 500 ml × (0.001 liter/ ml) = 0.50 liter.
Number of moles = concentration × volume = (0.0032 mole/ liter) × (0.50 liter) = 0.0016 mole.
Mass of KOH = number of moles × molar mass = 0.0016 mole × (56.11 g/ mole) = 0.09 g.

Note: potassium hydroxide is a hygroscopic material (it picks up atmospheric moisture). Its moisture content can be greater than 15% in some cases. This moisture content should be accounted for when preparing solutions. Accurate concentration of prepared potassium hydroxide solutions can be obtained by titration with a strong acid (e.g. hydrochloric acid).

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Calculate mass of ammonium chloride needed to prepare 1000 ml solution with pH value equal to 6.2.

General information:
Ammonium chloride (NH₄Cl) is a salt of ammonia (weak base). Ammonium chloride completely ionizes in solution to produce ammonim (NH₄⁺) and chloride (Cl^-) ions as shown in the following formula:

NH₄Cl_{(in aqueous solution)} à NH₄⁺_{(in aqueous solution)} + Cl^-_{(in aqueous solution)} 

Ammonium ions act as weak monoprotic acid in solution. It partially ionizes (dissociates) to produce ammonia (NH₃) and hydronium ions (H₃O⁺) as shown in the following formula:

NH₄⁺ + H₂O ↔ NH₃ + H₃O⁺

Ammonium chloride has a molar mass (molecular weight) equal to 53.49 g/ mole. Ammonia has Kb value equal to 1.74 × 10^-5  at 25 ºC.

Molar concentration of a weak monoprotic acid required to produce certain pH value in solution can be calculated from the following equations (see where it came from):

[H₃O⁺] = [H⁺]
C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.

Calculations:
Step 1.  calculate Ka value of ammonium chloride from Kb value of ammonia as the following:
Ka × Kb = Kw
Ka = Kw/ Kb
Where Kw is the dissociation constant of water.
Ka = (1.00 × 10^-14)/ (1.74 × 10^-5)
Ka = 5.75 × 10^-10

Step 2. Calculate molar concentration required to produce solution pH equal to 3.2 using the above-mentioned equations. 
pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-6.2 = 6.31 × 10^-7 mole/ liter

C = ((2 × 6.31 × 10^-7 + (5.75 × 10^-10))² - (5.75 × 10^-10)² ) ÷ (4 × (5.75 × 10^-10))

C = ((1.26 × 10^-6)² - (3.31 × 10^-19 )) ÷ (2.30 × 10^-9)

C = ((1.59 × 10^-12) - (3.31 × 10^-19)) ÷ (2.30 × 10^-9)

C= (1.59 × 10^-12) ÷ (2.30 × 10^-9)

C = 0.00069 mole/ liter

Use online pH calculator to check that 0.00069 mole/ liter ammonium chloride produces solution pH = 6.2. 

Step 3. Determine mass of ammonium chloride needed to prepare the solution.
Volume of solution in liters = 1000 ml × (0.001 liter/ ml) = 1.0 liter
Number of moles = concentration × volume = (0.00069 mole/ liter) × 1.0 liter = 0.00069 mole

Note: one mole ammonium chloride produces one mole ammonium ions in solution. Therefore, 0.00069 mole ammonium ions are produced from 0.00069 mole ammonium chloride.

Mass of ammonium chloride = number of moles × molar mass =0.00069 mole × (53.49 g/ mole) = 0.037 g

To make the preparation, dissolve 0.037 g ammonium chloride in freshly distilled water to make 1000 ml solution (using 1000 ml volumetric flask).

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Calculate mass of salicylic acid needed to prepare 500 ml solution with pH equal to 2.7.

General information:
Salicylic acid is a weak monoprotic acid. It partially ionizes (dissociates) in solution to produce hydronium and salicylate ions. It has a molar mass (molecular weight) equal to 138.12 g/ mole. It has Ka value equal to 1.06 × 10^-3  at 25 ºC.

Molar concentration of a weak monoprotic acid required to produce certain pH value in solution can be calculated from the following equations (see where it came from):

[H₃O⁺] = [H⁺]
C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.


Calculations:

pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-2.7 = 0.0020 mole/ liter

C = ((2 × 0.0020 + (1.06 × 10^-3))² - (1.06 × 10^-3)² ) ÷ (4 × (1.06 × 10^-3))

C = ((0.00506)² - (1.12 × 10^-6 )) ÷ (0.00424)

C = ((2.56 × 10^-5) - (1.12 × 10^-6 )) ÷ (0.00424)

C= (2.45 × 10^-5) ÷ (0.00424)

C = 0.00578 mole/ liter

Use online pH calculator to check that 0.00578 mole/ liter salicylic acid produces solution pH = 2.7. 

Mass of salicylic acid needed to make 500 ml solution with a concentration of 0.00578 mole/ liter is calculated as the following:

Volume of solution in liters = 500 ml × (0.001 liter/ ml) = 0.5 liter
Number of moles = concentration × volume = (0.00578 mole/ liter) × 0.5 liter = 0.00289 mole

Mass of salicylic acid = number of moles × molar mass = 0.00289 mole × (138.12 g/ mole) = 0.40 g

To make the preparation, dissolve 0.40 g salicylic acid in freshly distilled water to make 500 ml solution (using 500 ml volumetric flask).

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Prepare 500 ml acetic acid solution with a pH value equal to 2.7.
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Prepare 500 ml acetic acid solution with a pH value equal to 2.7

General information:
Acetic acid is a monoprotic weak acid. It partially ionizes (dissociates) in solution to produce hydronium and acetate ions. It has a molar mass (molecular weight) equal to 60.05 g/ mole. Ka value for acetic acid at 25 ºC is equal to 1.75 × 10^-5.

Molar concentration of acetic acid solution required to produce certain pH value in solution can be calculated from the following equations:

[H₃O⁺] = [H⁺]
C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.


Calculations:

pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-2.7 = 0.0020 mole/ liter

C = ((2 × 0.0020 + (1.75 × 10^-5))² - (1.75 × 10^-5)² ) ÷ (4 × (1.75 × 10^-5))

C = ((0.004)² - (3.06 × 10^-10 )) ÷ (0.00007)

C = ((1.6 × 10^-5) - (3.06 × 10^-10 )) ÷ (0.00007)

C= (1.6 × 10^-5) ÷ (0.00007)

C = 0.229 mole/ liter

Use online pH calculator to check that 0.229 mole/ liter acetic acid produces solution pH = 2.7 

Mass of acetic acid needed to make 500 ml solution with a concentration of 0.229 mole/ liter is calculated as the following:

Volume of solution in liters = 500 ml × (0.001 liter/ ml) = 0.5 liter
Number of moles = concentration × volume = (0.229 mole/ liter) × 0.5 liter = 0.115 mole

Mass of acetic acid = number of moles × molar mass =0.115 mole × (60.05 g/ mole) = 6.91 g

To make the preparation, dissolve 6.91 g acetic acid in freshly distilled water to make 500 ml solution (using a volumetric flask).

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Calculate mass of acetic acid needed to prepare 100 ml solution with pH equal to 3.5

General information:
Acetic acid is a monoprotic weak acid. It partially ionizes (dissociates) in solution to produce hydronium and acetate ions. It has a molar mass (molecular weight) equal to 60.05 g/ mole. Ka value for acetic acid at 25 ºC is equal to 1.75 × 10^-5.

Molar concentration of acetic acid solution required to produce certain pH value in solution can be calculated from the following equations:

[H₃O⁺] = [H⁺]

[H⁺] = (- Ka + sqrt(Ka² + 4 × Ka × C)) ÷ 2

Rearranging the above equation gives the following:

C = ((2 × [H⁺] + Ka)² - Ka² ) ÷ (4 × Ka)

Where C is molar concentration of the monoprotic acid, [H⁺] is molar concentration of hydrogen (hydronium) ion, and Ka is the value of the acid ionization constant.


Calculations:

pH = - log [H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-pH = 10^-3.5 = 0.000316 mole/ liter

C = ((2 × 0.000316 + (1.75 × 10^-5))² - (1.75 × 10^-5)² ) ÷ (4 × (1.75 × 10^-5))

C = ((0.00065)² - (3.0625 × 10^-10 )) ÷ (0.00007)

C = ((4.225 × 10^-7) - (3.0625 × 10^-10 )) ÷ (0.00007)

C= (4.222 × 10^-7) ÷ (0.00007)

C = 0.006 mole/ liter

Use online pH calculator to check that 0.006 mole/ liter acetic acid produces solution pH = 3.5 

Mass of acetic acid needed to make 100 ml solution with a concentration of 0.006 mole/ liter is calculated as the following:

Volume of solution in liters = 100 ml × (0.001 liter/ ml) = 0.1 liter
Number of moles = concentration × volume = (0.006 mole/ liter) × 0.1 liter = 0.0006 mole

Mass of acetic acid = number of moles × molar mass =0.0006 mole × (60.05 g/ mole) = 0.036 g

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Calculate mass of hydrochloric acid (HCl) needed to prepare 100 ml solution with pH equal to 1.5

General information:
HCl is a strong acid. It completely ionizes (dissociates) in solution to produce hydronium and chloride ions. It has a molar mass (molecular weight) equal to 36.46 g/ mole.

Calculations:
Molar concentration for HCl solution with pH equal to 1.5 is calculated as the following:

pH = -log[H₃O⁺]
[H₃O⁺] = 10^-pH

[HCl] = [H₃O⁺] = 10^-1.5 = 0.032 mole/ liter

Number of moles and mass of HCl needed to make 100 ml solution with a concentration of 0.032 mole/ liter are calculated as the following:

Volume of solution in liters = 100 ml × (0.001 liter/ ml) = 0.1 liter
Number of moles = concentration × volume = (0.032 mole/ liter) × 0.1 liter = 0.0032 mole

Mass of HCl = number of moles × molar mass = 0.0032 × (36.46 g/ mole) = 0.12 g

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Calculate pH of 100 ml solution containing 0.05 g acetic acid and 0.06 g sodium acetate.

General information:
Sodium acetate completely ionizes in solution to produce sodium ions and acetate ions according to the following formula:

NaAcetate_{(in aqueous solution)} → Na⁺_(aq) + Acetate^-_(aq)

Accordingly, the number of moles of acetate ions in solution is equal to that of  sodium acetate. Acetic acid and sodium acetate (acid - conjugare base) in solution compose a buffer system. The pH can be calculated using the buffer equation or using the pH equation for a mixture of an acid with its conjugate base. Molar concentrations must be calculated before using any of these equations.

Molar mass (molecular weight) of acetic acid = 60.05 g/ mole
Molar mass (molecular weight) of sodium acetate = 82.03 g/ mole
Acetic acid has Ka = 1.75 × 10^-5 and pKa = 4.76

Calculations:
Volume of solution in liters = 100 ml × (0.001 liter/ ml) = 0.1 liter
Number of moles of acetic acid in solution = 0.05 g ÷ (60.05 g/ mole) = 0.0008 mole
Molar concentration of acetic acid in solution = 0.0008 mole ÷ (0.1 liter) = 0.008 M
Number of moles of sodium acetate in solution = 0.06 g ÷ (82.03 g/ mole) = 0.0007 mole
Molar concentration of sodium acetate in solution = 0.0007 mole ÷ (0.1 liter) = 0.007 M

Using the buffer equation:
pH = pKa + log ([acetate]/[acetic acid])
pH = 4.76 +  log (0.007/0.008) = 4.76 - 0.06 =4.70

The same answer is obtained using the pH equation for a mixture of an acid with its conjugate base or simply using online pH calculator.

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