Question

using standardized value of NaOH (.092093 M) and .2527 grams of malonic acid and the 70...

using standardized value of NaOH (.092093 M) and .2527 grams of malonic acid and the 70 mL to get to the firts end point. calculate the MW of malonic acid

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1. A sample of hydrochloric acid is standardized using sodium bicarbonate. a. Write the balanced chemical...
1. A sample of hydrochloric acid is standardized using sodium bicarbonate. a. Write the balanced chemical equation for the reaction of hydrochloric acid with sodium bicarbonate, NaHCO3 b. Calculate the molar concentration of the hydrochloric acid if 35.18 mL of hydrochloric acid was required to neutralize 0.450 g of sodium bicarbonate to a phenolphthalein end point. c. Calculate the mass percent concentration of hydrochloric acid using the molar concentration calculated in part b and assuming the density of the solution...
Make 50.0 mL of a solution that is both 0.15 M in malonic acid, CH2(CO2H)2 and...
Make 50.0 mL of a solution that is both 0.15 M in malonic acid, CH2(CO2H)2 and 0.020 M in MnSO4. Calculate how many grams of malonic acid to make 50.0 ML of 0.15 M malonic acid an dhow many grams of MnSO4 + H2O needed to make 50.0 mL of 0.020 M manganese(II) sulfate solution. Note for a hydrate, the water molecule in the formula is part of what is weighed out and must be included in the molar mass...
If 1.20 grams of impure solid KHP sample required 2.53 mL of your standardized NaOH to...
If 1.20 grams of impure solid KHP sample required 2.53 mL of your standardized NaOH to reach the end point, what was the percent KHP in this sample? mm= 204.2 NaOH M= 0.127 M
A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with 0.100 NaOH....
A 25.00 mL aliquot of 0.100 F malonic acid (propanedioic acid) is titrated with 0.100 NaOH. a) the initial pH (before addition of any NaOH) is? b)What is the pH at the first half-equivalence point (halfway to the first equivalence point)? c)What is the pH at the first equivalence point? d)What is the pH at the second half-equivalence (halfway between the first and second equivalences)? e)What is the pH at the second equivalence point? f)What is the pH after adding...
A NaOH solution is standardized using KHP and the molarity of the NaOH solution is determined...
A NaOH solution is standardized using KHP and the molarity of the NaOH solution is determined to be 0.4150 M. Since the acid in vinegar is the monoprotic acid, acetic acid, the concentration of the acetic acid can easily be determined by titration. If 90.72 mL of this solution is required to titrate 10.32 mL of vinegar to the Phenolphthalein endpoint, what is the concentration of acetic acid in the vinegar? _______________ M 95.71 mL of NaOH solution is required...
You are titrating 50mL of diprotic acid H2X with standardized 0.01M NaOH solution. It takes 25...
You are titrating 50mL of diprotic acid H2X with standardized 0.01M NaOH solution. It takes 25 mL of NaOH to reach the first equivalence point. a. Write down the balanced equation for each of the titration reactions b. What is the volume of base needed to reach the second equivalence point c. Calculate the concentration (M) of the unknown acid solution, H2X
A standardized 0.1000 M NaOH solution was used to determine the amount of H3PO4 and KH2PO4...
A standardized 0.1000 M NaOH solution was used to determine the amount of H3PO4 and KH2PO4 in a 25.00-mL unknown sample. It's known that 11.05 mL of the NaOH solution was consumed to reach the first equivalence point and an additional 21.60 mL of the NaOH solution was used to reach the second end point. a) Please determine the concentration of the H3PO4 present in the unknown. b) What's the concentration of the KH2PO4 present in the unknown?
A 50.00 mL sample of a citrus drink requires 17.62 mL of 0.04166 M NaOH to...
A 50.00 mL sample of a citrus drink requires 17.62 mL of 0.04166 M NaOH to reach the phenolphthalein end point (pH ≈ 9). Express the sample’s acidity as grams of citric acid, C6H8O7, per 100 mL. Hint: citric acid is a polyprotic acid. You need to determine first what is the end point of the titration from the three pKa (3.128, 4.761, 6.396) and end point pH in order to get the correct acid-base reaction.
A titration is done using a 0.1302 M NaOH solution to determine the molar mass of...
A titration is done using a 0.1302 M NaOH solution to determine the molar mass of a monoprotic weak acid (HA). If 50.00 mL of a 1.863 g sample of the unknown acid (HA) is titrated to the equivalence point/end point with 70.11 mL of a 0.1302 M NaOH aqueous solution, what is the molar mass of the unknown acid?
1b. Calculate the exact molarity of a solution of NaOH if 55.00 mL of it is...
1b. Calculate the exact molarity of a solution of NaOH if 55.00 mL of it is needed to titrate an amount of KHP that equals your average mass (that you stated above). Average is 50.0 (4 pts) 2a. What is the average molarity for your standardized NaOH solution?_______________ (1 pt) Average: .0075 M 2b. If 1.20 grams of impure solid KHP sample required 2.53 mL of your standardized NaOH to reach the end point, what was the percent KHP in...