a solution is prepared by dissolving a 90.0 g of benzoIc acid (HC7H5O2) and 90.0
ID: 877996 • Letter: A
Question
a solution is prepared by dissolving a 90.0 g of benzoIc acid (HC7H5O2) and 90.0 g of sodium benzoate (NaC7H5O2) in enough water to give total volume of 500 ml
a. calculate the pH of the solution?
b. A 10.0 ml sample of 0.50 M HCl is added to 90.0 ml of the solution from part (a). assuming volumes are additve, calculate the pH of the solution.?
C. you are tasked with making exactly 5.00L of a pH 4.50 buffer using 1.0 M benzoic acid and 1.5 M sodium benzoate solutions. what volume of each solution will you need to use to prepare this buffer.?
Explanation / Answer
a solution is prepared by dissolving a 90.0 g of benzoIc acid (HC7H5O2) and 90.0 g of sodium benzoate (NaC7H5O2) in enough water to give total volume of 500 ml
Solution:
Given: Mass of benzoic acid =90.0 g , sodium benzoate = 90.0 g
Total volume = 500 mL = 0.500 L
We find moles of benzoic acid and sodium benzoate and then concentration.
Mol Benzoic acid = Mass in g / molar mass of benzoic acid
= 90.0 g / 122.12 g /mol = 0.737 mol
Mol sodium benzoate = 90.0 g / 144.11 g per mol = 0.624 mol
[Benzoic acid] = 0.737 mol / 0.500 L = 1.474 M
[Sodium benzoate ]= 0.624 mol / 0.500 = 1.249 M
we use Henderson equation to calculate pH
Ka of benzoic acid = 6.46 E -5
Pka = - log ka = -log ( 6.46E-5) = 4.1898
Henderson equation
pH = pka + log ([sodium benzoate ] / [ benzoic acid ])
= 4.1898 + log ([1.249]/[ 1.474 ]
= 4.1898 -0.0719
= 4.12
a. calculate the pH of the solution?
b. A 10.0 ml sample of 0.50 M HCl is added to 90.0 ml of the solution from part (a). assuming volumes are additve, calculate the pH of the solution.?
Sodium benzoate will react with HCl and forms benzoic acid again
CH3COONa (aq) + HCl --- > CH3COOH (aq) + NaCl (aq)
From this reaction the mol ratio of CH3COONa to HCl to CH3COOH is 1 : 1 : 1
We can say that 1 mol of HCl forms one mol of CH3COOH and decrease 1 mol sodium benzoate.
mol of HCl = volume in L * molarity = 0.010 L * 0.50 M= 0.005 mol
Final moles of benzoic acid = 1.474 + 0.005 = 1.479 mol
[Benzoic acid ]= 1.479 mol / ( 0.50 + 0.010 ) L
= 2.90 M
Final moles of sodium benzoate = 1.249 – 0.005 mol = 1.244 mol
[Sodium benzoate ] = 1.244/ 0.510L = 2.44 M
pH = 4.1898 + log ([2.44/2.90 )
pH = 4.1898 – 0.07514 = 4.11
C. you are tasked with making exactly 5.00L of a pH 4.50 buffer using 1.0 M benzoic acid and 1.5 M sodium benzoate solutions. what volume of each solution will you need to use to prepare this buffer.
Solution:
Volume of benzoic acid = x
Volume of sodium benzoate = y
x + y = 5.00 ……….1
we find ratio of sodium benzoate : benzoic acid by using Henderson equation
pH = 4.1898 + log ([Sodium benzoate ] / [benzoic acid ] )
4.50 = 4.1898 + log ([Sodium benzoate ] / [benzoic acid ])
log ([Sodium benzoate ] / [benzoic acid ]) = 0.3102
Lets take antilog of both side
([Sodium benzoate ] / [benzoic acid ] )= Antilog 0.3102
([Sodium benzoate ] / [benzoic acid ]) = 2.04
Now we have assumed volume of sodium benzoate = y and volume of benzoic acid = x
We calculate moles of benzoic acid and sodium benzoate
Moles of benzoic acid = 5.0 L * 1.0 M = 5 mol benzoic acid
Moles of sodium benzoate = 5.0 L * 1.5 = 7.5 mol sodium benzoate
In above ratio we use the moles of both
(Mole sodium benzoate /volume of sodium benzoate in L)/(moles of benzoic acid / volume of benzoic acid )
= 2.04
( 7.5 mol / vol. of sodium benzoate ) / ( 5.0 mol / vol. of benzoic acid) = 2.04
( 7.5 mol / y ) / ( 5.0 mol /x ) = 2.04
(7.5 / y ) / (5.0/ x ) = 2.04
7.5 x/ 5.0 y = 2.04
x/ y = 1.36
x = 1.36 y
lets plug this value in equation 1
1.36 y + y = 5.0 L
2.36 y = 5.0 L
y = 2.12 L
Volume of sodium benzoate = 2.12
Volume of benzoic acid = 5.0 – 2.12 = 2.88 L
So buffer is prepared by addition of volume of 1.0 M benzoic acid is 2.88 L
And sodium benzoate of 1.5 M volume = 2.12 L
If we mix above two solutions then we get desired buffer.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.