1J /(g*C), 1*C, 1 J, m=g
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Answer :
The mass of sodium benzoate that can be added to benzoic acid is 5.07 g
Explanation :
The problem can be solved with the help of Henderson-Hasselbalch equation.
Step 1 : Find concentration of sodium benzoate
The equation is given below.
We have been given,
pH = 4.30
[Acid] = 0.17 M
Ka of benzoic acid can be found out using standard reference table which is 6.5 x 10⁻⁵
pKa = - log (Ka)
pKa = - log (6.5 x 10⁻⁵)
pKa = 4.19
Let us plug in the above values in Henderson equation.
The functions cancel out each other.
[Base] = 0.22 M
The base in this case is sodium benzoate.
Therefore concentration of sodium benzoate is 0.22 M.
Step 2 : Find moles of sodium benzoate
The volume of the solution is 160 mL.
Volume in L =
Moles of sodium benzoate can be found as,
Moles of sodium benzoate are 0.0352
Step 3 : Find mass of sodium benzoate
Mass of sodium benzoate can be calculated as
Mass in grams = Moles x Molar mass
The formula of sodium benzoate is C₇H₅O₂Na.
Molar mass of sodium benzoate = 7 x 12.01 + 5 x 1.01 + 2 x 16 + 22.98
Molar mass of sodium benzoate = 144.1
Mass of sodium benzoate =
Mass of sodium benzoate = 5.07 g
5.07 grams of sodium benzoate should be added to the given volume of benzoic acid.