Does anyone mind helping me out with this problem? I know how to find the averag
ID: 1047650 • Letter: D
Question
Does anyone mind helping me out with this problem? I know how to find the average but not the other calculations required. Thank you! CHM 147 Procedure A. Standardization of NaOH Solution Use a small graduated cylinder to measure about 10mL of stock NaOH solution. Combine the 10 mL of NaOH solution with about 400 mL of deionized water in a 500 mL Erlenmeyer flask. Stopper the flask and mix thoroughly over a period of 5 minutes before using. This is all the NaOH solution you will have for the entire experiment; do not waste it! Clean the buret, rinse with distilled water, and then rinse with a few mL of the NaOH solution you have prepared. Before filling buret. Fill the buret with the prepared NaOH solution and open the stopcock to fill the tip. Drain the buret to a volume between 0-5 mL mark. Remember you should not start at 0.00 mL. Check to see that there are no air bubbles in the buret or the tip Weigh 0.5XX grams of potassium hydrogen phthalate (KHC,H.0.) on the analytical balance and place in a 250mL Erlenmeyer flask. Dissolve the solid in 50mL of distilled water, add 2-3 drops of phenolphthalein indicator, and the magnetic stir bar. Place the flask on the magnetic stir plate with a white sheet of paper under the flask to aid in detection of any color change. Record the initial volume in the buret. Add the NaOH from the buret to the acid in the flask. Level of meniseus A sheet of white paper or towel below the fask w help in recognising the calor change st the endpoint Swirl the fark contn uously unsl ane drep af titrant cavies a color change throughout the entire solution Mix the solution using a magnetic stirrer or swirl the flask continuously as you a (Stir the solution at a reasonahle rate hut not too ui the NaOH irl the flask continuously as you add the NaOH.Explanation / Answer
Molar mass of Potassium Hydrogen phthalate = 204.22gm/mol
Moles of Potassium Hydrogen phthalate used = 0.5gm/ 204.22gm/mol = 2.45 *10^-3 moles
Molarity of the solution = 2.45 *10^-3 moles *1000mL/50mL = 0.049 M
Calculate the strength of NaOH by using this molarity.
Trial 1 :
M1V1 = M2V2
where M1 and V1 is the molarity and volume of NaOH added. m2 and v2 is the molarity and volume of potassium hydrogen phthlate.
M1 = M2V2/V1 = 0.019 M *50mL/17.56 mL = 0.14 M
Molarity of NaOH = 0.14 M
At the neutralization point number of moles of base = number of moles of acid
Moles of NaOH added = 0.14 *25.85 mL/1000mL = 3.62 *10^-3 moles
moles of acid = 3.62 *10^-3
mole= mass/molar mass of acid
molar mass of acid = 0.250gm/ 3.62 *10^-3 mol = 69.06gm/mol
The unknown acid is Acetic acid.
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Total volume of acid + deionized water = 5 mL + 25 mL = 30mL
m1V1 = M2V2
where M1 and V1 is the molarity and volume of acetic acid . m2 and V2 is the molarity and volume of NaOH
M1 = M2V2/V1 = 29.78 mL * 0.14 M/30mL = 0.138 M
The vinegar is 0.138M in acetic acid concentration.
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