Question

1. How many grams of pure, solid NaOH are required to make 400 mL of 0.1M NaOH solution? 2. How many grams of a solution that is 50% by weight NaOH is required to make 400 mL of 0.1M NaOH solution? 3. The density of a 50% solution of NaOH is 1.525 g/mL. What volume of a solution that is 50% by weight NaOH is required to make 300 mL of 0.1M NaOH solution? 4. Why does weighing by difference eliminate systematic balance errors? 5. For each of your 3 acceptable trials (not the cursory one) in standardizing your NaOH against KHP, what are the mass in grams of KHP for each of the three samples to one thousandth of a gram, e.g. 0.507g? 6. What is the volume in mL of NaOH required to reach the endpoint for each of the KHP samples and enter the volume to a hundredth of a ml, e.g. 15.24 mL? 7. Calculate the values for each of your three trials for the molarity of your NaOH solution. Place your molarity entries in the order corresponding with the masses of KHP and the volumes of NaOH required for the trials. 8. Calculate the value for the average of the three trials determining the molarity of your NaOH solution. 9. Calculate the value for the standard deviation of the average molarity 10. Calculate the value for the 90% confidence limits for the molarity of your NaOH solution. 11. Why does it not matter how much water you added when dissolving the acid or when carrying out the titration? 12. What was the brand name of your sample commercial vinegar solution? 13. What volume in mL did you use for each of the three acceptable trial samples of your vinegar analysis? 14. What was the volume of standardized NaOH solution that was required for each trial determining the amount of acetic acid in your commercial vinegar sample?

Answer 1

2 grams of Mouth = Molanity & volume x Molecular weight of nown ② Given, 50% means 50g/100 me → 50g in 100 me - 500g in loomL (IL) I CinL) 0.X 4LX 409lmal 2009 1.525 gione density = deney 0 grame of mooh - 2oog in toome 50%. solution = 1.525 gimL 50% by weight Moon = 200g 50% buon Volume Mud 2004 HAL • difference is a very accurate method which detlemines mars by difference between the two readnigs i mars au thies systematic euol & in the absolute man could be elminated by substraction of final weight from the initial wegght. 6 You can determine volume ime of Haoh required to reach the end point by : Vol (ML) = molarity of Khp x volume of Kulp MADH Molan'ty of Maori o some of your questious would be done when you had are done with your trials and the values are obtained on adding water dissolving the acid would not affect Aitration since this would only affect Hoot buc became increase in volume with hame number of moles decreases the concentration of analyte, thus increares the pH of solution. But since you are wring an indicator to determine the titration and to reach the end pouit hence, no enor in titration until indicator conc. is changed . dwling Atration score in volumen of analyle wring an indicad