The general reaction of oxalic acid and NaOH is as follows:
H2C2O4 (aq) + 2 NaOH = Na2C2O4 (aq) + 2 H2O(l)
Number of mole s= molarity * volume in L
= 0.538 mole / L * 29.78 ml *1 L /1000 ML
= 0.01602 Moles NaOH
Now calculate the moles of acid as follows:
0.01602 Moles NaOH*1 mole H2C2O4 /2 Moles NaOH
= 0.00801 Moles H2C2O4
Molarity = number of moles / volume in L
= 0.00801 Moles H2C2O4/14.75ml /1 L/1000 ml
= 0.543 M or mole s/ L
Oxalic acid, H2C204, is used in the restoration of old wood. What is the molarity of...
6. Oxalic acid H2C2O4 is a diprotic acid and can be used, in dilute form, as wood bleach. pKa1 = 1.237 and pKa2=4.187 A. How many milliliters of 0.0500 M NaOH would be required to completely neutralize 10.0 mL of 0.100 M H2C2O4? (5 pts) B. Estimate the pH at the first equivalence point of the titration. (6 pts) C. What is the pH at the second equivalence point of the titration. (8 pts) Please help with work shown, especially...
The reaction of MnO4 with oxalic acid (H2C204) in acidic solution, yielding Mn2+ and CO2 gas, is widely used to determine the concentration of permanganate solutions. a. Write a balanced equation for the reaction. b. Show that the reaction goes to completion by calculating the values of AGº and K at 25 °C. Most values for AG°f are in the back of your textbook. AG°f of H2C2O4(aq) is -674.0 kJ/mol. If your calculator will not give you an actual value...
Data Table 1 Mass of flask and oxalic acid (g) 117.43 Mass of empty flask (g) 116.93 Mass of oxalic acid (g) 0.5 Moles of oxalic acid (mol) Final volume of NaOH (mL) 17 Initial volume of NaOH (mL) 5 Volume of NaOH used (mL) 12 Moles of NaOH (mol) Molarity of NaOH (M) Data Table 2 Mass of flask and vinegar (g) 126.61 Mass of empty flask (g) 121.63 Mass of vinegar (g) 4.98 Final volume of NaOH (mL)...
Oxalic acid can be used, in dilute form, as wood bleach. How many milliliters of 0.050 M NaOH would be required to neutralize 10.0 mL of 0.100 M H2C2O4? Estimate the pH at the first equivalence point of the titration. This question! Estimate the pH at the second equivalence point of the titration.
please can you help me Consider the following two reactions involving oxalic acid. 2Mn04- + 5H2C204 + 6H+ + 2Mn2+ + 10002 + 8H20 H2C2O4 + 204- → C2042- + 2H20 What volume of 0.0100 M KMnO4 solution will titrate a solution prepared from 37.0 mg of H2C204? 2.67 mg Submit Answer Incompatible units. No conversion found between "mg" and the required units. Tries 0/5 Previous Tries What volume of 0.0100 M NaOH solution will titrate a solution prepared from...
moles of oxalix acid moles/molarity of NaOH average molarity volume/molarity of hcl used average molarity NAME SECTION DATA SHEET LOCKER A CTOR Write a balanced equation for the reaction between H.CO. and NaOH: 90.33/mol 0.878 0836 TRIAL NUMBER a. Mass of oxalic acid used (g) b. Final Buret reading (ml.] c. Initial Buret reading (ml] d. Volume of NaOH used (b-c) Moles of oxalic acid used f. Moles of NaOH used (2 xe) g. Molarity of NaOH (f/d/1000) Average Molarity...
The empirical formula of the product is (1 pt) Consider the following reaction between oxalic acid and potassium hydroxide. 2. H20 + K2C204 H2C204 + KOH A 1.500 g sample of oxalic acid is dissolved in water and titrated with 32.67 mL of KOH to a phenolphthalein end point. What is the molar concentration of the KOH solution? (5 pts)
wote TITRATION OF AN ACID WITH A BASE 2020 Pre-lab assignment ame In part A of the procedure, you will make a 0.2 M NaOH solution by diluting 6 M NaOH. How many milliliters of 6 M NaOH are required to make 500.0 mL of 0.2 M NaOH? Even though it will be breaking significant figure rules, give the answer to the nearest tenth. In part B of the procedure you will be dissolving oxalic acid dihydrate in water to...
Trial one Trial two Trial three Volume of oxalic acid (mL) 10.1 mL 10.2 mL 10.1 mL Moles of oxalic acid (moles) 0.0026 0.0026 0.0026 Initial volume of NaOH (mL) 18.7 mL 26.5 mL 34.6 mL Final volume of NaOH (mL) 26.5 mL 34.6 mL 42.6 mL Delivered volume of NaOH (moles) 7.8 mL 8.1 mL 8 mL Moles of NaOH (moles) 0.0052 0.0052 0.0052 Molarity of NaOH (M) 0.6667 0.6420 0.65 Average Molarity NaOH (M) 0.65 Concentration oxalic acid...
Questions 1. Calculate the molarity of a sodium hydroxide (NaOH) solution that is titrated with 0.6887 g of oxalic acid (Equation 2). The titration requires 15.80 mL of the NaOH solution to reach the end point. Calculate the molarity of a sulfuric acid (H,SO) solution if 30.10 mL of 0.62 10 M NaOH is required to reach the end point when titrated against 10.00 mL of the unknown acid solution. The balanced chemical equation for the reaction is given below....