The Ka of benzoic acid is 6.5 x 10 ^(-5) a) Calculate the pH of a 40.00 mL, 0.1

Question

The Ka of benzoic acid is 6.5 x 10 ^(-5)

a) Calculate the pH of a 40.00 mL, 0.1 M benzoic acid buffer solution after the addition of 20.00 mL of a 0.1M NaOH

b) Calculate the pH of a 40.00 mL, 0.1 M benzoic acid buffer solution after the addition of 50.00 mL of a 0.1M NaOH

REDOX TITRATION

2) Express the reactions for potassium permanganate titration with sodium oxalate: Hint: include the 2 half-reactions.

3) What is the percent oxalate in a 1.723 g sample if 31.10 mL of 0.019M permanganate solution is required for titration?  

Explanation / Answer

moles of benzoic acid in 40ml of 0.1M= 0.1*40/1000=0.004, moles of NaOH in 20ml of 0.1M NaOH =0.1*20/1000 =0.002

the reaction between benzoic acid and sodium hydroxide is C6H5COOH+NaOH ----->C6H5COONa+H2O

According to the reaction. 1 moles of benzoic acid and 1mol of sodium hydroxide. In the present case, excess is benzoic acid. Excess benzoic acid =0.004-0.002 =0,002 moles

moles of sodium benzoate =0.002 ( suppliments C6H5COO- )

volume after mixing = 40+20= 60ml =60/1000

pH= pKa + log [conjugate base/ acid]

Ka= 6.5*10-5, pKa= 4.187 , pH= 4.187+ log {(0.002*1000/60/ 0.002*1000/60} = 4.187

b) moles of sodium hydroxide in 50ml of 0.1M= 0.1*50/1000 =0.005

In this case, NaOH is excess, and excess moles =0.005-0.004=0.001

all the benzoic acid is neutralized. NaOH is strong base and ionizes completely.

volume of mixer = 40+50= 90ml =90/1000=0.09L

concentration of NaOH= 0.001/0.09=0.011

pOH= -log (0.011)=1.95

pH= 14-1.95= 12.05

2)

C2O42-         2 CO2 + 2 e- - Oxidation (1)

8 H+   +   MnO4-   +   5 e-          Mn2+   +   4 H2O- Reduction (2)

mutliply Eq.1 with 5 and Eq.2 with 2 to tive and then add to give

16 H+   +   2 MnO4-   +    5 C2O42-         2 Mn2+   +   10 CO2    +   8 H2O

moles of permanganate in 31.1 ml of 0.019 =0.019*31/1000=0.000589

2 moles of KMnO4 gives 5 moles of sodium oxalate

0.000589 moles of KMnO2 gives 5*0.000589/2= 0.001473

molar mass of sodium oxalate =134, mass of oxalate =molar mass* moles = 134*0.001473=0.1973gm

percent oxalate =100*0.1973/1.723=11.45%

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