# What is the average concentration of NaOH solution provided the average mL of NaOH added in 3…

What is the average concentration of NaOH solution provided the average mL of NaOH added in 3 titrations? I have provided some of my previous answers for reference. Tha
2. Amount of benzoic acid to be neutralized by 20 mL NaOH solution, in both moles and grams: NaOH [M]= 0.020 L*0.494 M= 0.00988 M Benzoic acid [M] = 0.00988 moles Benzoic acid (g)= 0.00988 mol*122.12 g/mol = 1.207 g
3. Actual mL values of NaOH solution delivered in 3 titrations:
Titration #1 20.07 mL Titration #2 20.07 mL Titration #3 2
What is the average concentration of NaOH solution provided the average mL of NaOH added in 3 titrations? I have provided some of my previous answers for reference. Tha
2. Amount of benzoic acid to be neutralized by 20 mL NaOH solution, in both moles and grams: NaOH [M]= 0.020 L*0.494 M= 0.00988 M Benzoic acid [M] = 0.00988 moles Benzoic acid (g)= 0.00988 mol*122.12 g/mol = 1.207 g
3. Actual mL values of NaOH solution delivered in 3 titrations: Titration #1 20.07 mL Titration #2 20.07 mL Titration #3 20.07 mL
4. Calculated average concentration of the NaOH solution, using the average mL amount of NaOH added in the 3 fine titrations:

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