Problem: Benzoic acid C6H5COOH is a weak acid with K a = 6.3 x 10-5C6H5CO2H (aq) + H2O (l) ⇌ H3O+ (aq) + C6H5CO2- (aq)a) Calculate the pH of a 0.150 M benzoic acid solution. Show all calulations.b) Suppose 1.44g of sodium benzoate, Na+C6H5COO-, is added to 100.0 mL of the 0.150 M benzioic acid solution. Calculate the pH of the resulting buffer solution assuming the volume of the solution does not change on addition of the solid.

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We’re being asked to calculate the pH of a) 0.150 M benzoic acid solution and b) the resulting buffer solution after adding 1.44 g sodium benzoate Na+C6H5COOto 100 mL of 0.150 M benzoic acid solution. 


Remember that weak acids partially dissociate in water and that acids donate H+ to the base (water in this case). 


The dissociation of C6H5CO2H is as follows:


C6H5CO2H (aq) + H2O (l) ⇌ H3O(aq) + C6H5CO2(aq)


For this problem, we need to do these steps:


Part  a:

Step 1Set up an ICE table to calculate the H3O+ ion concentration.

Step 2: Write an expression for the Ka to solve for concentration of H3O+

Step 3: Calculate the pH from H3O+ ion concentration


Part b:

Step 1: Calculate the moles of each species in the buffer solution 
Step 2Calculate the pH of the resulting buffer solution using the Henderson-Hasselbalch Equation:


pH=pKa+logconjugate baseweak acid

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Problem Details

Benzoic acid C6H5COOH is a weak acid with a = 6.3 x 10-5
C6H5CO2H (aq) + H2O (l) ⇌ H3O(aq) + C6H5CO2(aq)

a) Calculate the pH of a 0.150 M benzoic acid solution. Show all calulations.

b) Suppose 1.44g of sodium benzoate, Na+C6H5COO-, is added to 100.0 mL of the 0.150 M benzioic acid solution. Calculate the pH of the resulting buffer solution assuming the volume of the solution does not change on addition of the solid.

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