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'HC2H3O2' is a rather poor way of writing the formula for ethanoic(acetic) acid.
It is usually written as 'CH3COOH'.
lenpollock
Wiki User
∙ 13y agoNot a base. This is the condensed formula for acetic acid. The leading H is often telling.
CH3COOH is the more biologically influenced formula for acetic acid.
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What is the Conjugate base of HC2H3O2?
C2h3o2-
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?
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−]
HC2H3O2 is an electrolyte?
It is a weak electrolyte.
What is the gas evolved when magnesium react with HC2H3O2?
HC2H3O2 is acetic acid. Magnesium will react with water or acids to produce hydrogen gas.
How many valence electrons does HC2H3O2 have?
5 or 15
What is the Conjugate base of HC2H3O2?
C2h3o2-
Is vinegar a base in water?
The active ingredient in vinegar is acetic acid (HC2H3O2), which is a weak acid.
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?
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−]
Name of hc2h3o2?
Acetic acid
HC2H3O2 is an electrolyte?
It is a weak electrolyte.
What is the gas evolved when magnesium react with HC2H3O2?
HC2H3O2 is acetic acid. Magnesium will react with water or acids to produce hydrogen gas.
How do you neutralize HC2H3O2 plus LiOH?
Use equimolar quantities: LiOH + HC2H3O2 (acetic acid) --> C2H3O2- (acetate) + Li+ + H2O
What is dose hc2h3o2 stand for?
Acetic Acid
How many valence electrons does HC2H3O2 have?
5 or 15
How many moles are there in 500.0g of HC2H3O2 -?
8.32 MOLES
How many equivalent OH-1 are there in HC2H3O2?
5
How many number of carbon atoms are present in 0.062 mol acetic acid HC2H3O2?
The number of carbon atoms which are present in 0,062 mol acetic acid HC2H3O2 is 0,747.10e23.
