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Yes, NaC2H3O2 and HC2H3O2 is a buffer system when dissolved in water. Sodium acetate (NaC2H3O2) acts as a base, while acetic acid (HC2H3O2) acts as an acid. This buffer system can help maintain the pH of the solution when small amounts of acid or base are added.
There is only one equivalent of OH^-1 in one molecule of HC2H3O2.
Use equimolar quantities: LiOH + HC2H3O2 (acetic acid) --> C2H3O2- (acetate) + Li+ + H2O
The active ingredient in vinegar is acetic acid (HC2H3O2), which is a weak acid.
HC2H3O2, also known as acetic acid, is considered a weak electrolyte. This is because it partially dissociates into ions (H⁺ and C2H3O2⁻) when dissolved in water, but not completely like strong electrolytes do. Therefore, while it can conduct electricity to some extent, it does so less effectively than strong electrolytes.
The pH of a solution containing 0.1 M of HC2H3O2 is around 2.88.
Hydrochloric acid (HCl) is a strong acid, while acetic acid (HC2H3O2) is a weaker acid. In solution, HCl will dissociate completely into H+ and Cl- ions, while HC2H3O2 will only partially dissociate. This results in a higher concentration of H+ ions in HCl solution compared to HC2H3O2 solution at the same concentration.
To prepare 500 mL of a 0.5 M solution of acetic acid, you would need to take 100 mL of the 2.5 M stock solution and dilute it to 500 mL. This is because the molarity equation (M1V1 = M2V2) can be used to calculate the volume of stock solution needed for dilution.
Yes, acetic acid (HC2H3O2) is a weak electrolyte. It partially dissociates in water to form H+ and C2H3O2- ions, allowing for a small electrical conductivity in solution.
Yes, HCl will be fully ionised (1 mole gives 2 moles of ions) where as acetic acid will only partially be ionised, probably less than 1%, so 1 mole gives you less than 1.01 mole 'particles' in solution.
The conjugate base of HC2H3O2 is C2H3O2-. This ion is formed when HC2H3O2 donates a proton.
In aqueous HC2H3O2 solution, species present are acetic acid (HC2H3O2) molecules and hydronium ions (H3O+). The acetic acid molecules can partially dissociate to form acetate ions (C2H3O2-) and hydronium ions.
CHEM 123 Solution to Ques 16.65 from Wiley #12 16.65 A student added 100 mL of 0.10 M NaOH to 250 mL of a buffer that contained 0.15 M HC2H3O2 and 0.25 M C2H3O2−. What is the concentration of HC2H3O2 after the addition of the strong base. Begin by writing the equation for the buffer (acid dissociation of the weak acid) HC2H3O2 + H2O H3O + + C2H3O2- Ka = 1.8 × 10-5 Initial amounts in the solution are: # mol C2H3O2- = (0.25 mol/L)(0.25 L) = 0.063 mol C2H3O2- # mol HC2H3O2 = (0.15 mol/L)(0.25 L) = 0.038 mol HC2H3O2 The added base (0.100 L)(0.10 mol/L) = 0.010 mol NaOH) will react with the acetic acid present in the buffer solution: NaOH + HC2H3O2 ⎯→ H2O + NaC2H3O2 Assume the added base reacts completely (since it's a strong base with a weak acid). For each mole of base added, one mole of HC2H3O2 is converted to C2H3O2-. Since 0.010 mol of base is added, 0.010 mol of HC2H3O2 will be depleted from the initial amount (0.038 mol) and 0.010 mol C2H3O2− ADDITIONAL produced in this reaction (added to the initial amount of 0.063 mol C2H3O2−): # mol HC2H3O2final = (0.038 - 0.010) mol = 0.028 mol # mol C2H3O2-final = (0.063 + 0.010) mol = 0.073 mol The final volume of solution is 250 mL + 100 mL = 350 mL. AFTER the reaction with NaOH, the concentrations become… [HC2H3O2] = 0.028 mol/0.350L = 0.080 M HC2H3O2 [C2H3O2−] = 0.073 mol/0.350L = 0.21 M C2H3O2− In Wiley #12, the question asked only what the concentration of acetic acid becomes, so this would be the answer: 0.080 M. (The optional given you in Wiley was 0.079 M.) In the actual question in the book, it went further and asked you what is the CHANGE in concentration of HC2H3O2 and C2H3O2−, so you would simply go one further step: Initial conc of HC2H3O2 was 0.15M Change in conc = 0.15−0.080 = 0.0.07 M decrease in [HC2H3O2] Initial conc of C2H3O2− was 0.25M Change in conc = 0.25M − 0.21 Μ = 0.04 Μ decrease in [C2H3O2−]
To find the number of moles in 500.0g of HC2H3O2, divide the given mass (500.0g) by the molar mass of HC2H3O2 (60.05 g/mol). This calculation would give you approximately 8.33 moles of HC2H3O2 in 500.0g.
The main ingredient in vinegar is acetic acid (HC2H3O2), which is a weak acid.
That is the condensed formula for the weak acid acetic acid.CH3COOH is a common formula for acetic acid written by biochemists and biologists
Yes, NaC2H3O2 and HC2H3O2 is a buffer system when dissolved in water. Sodium acetate (NaC2H3O2) acts as a base, while acetic acid (HC2H3O2) acts as an acid. This buffer system can help maintain the pH of the solution when small amounts of acid or base are added.