at .2M and 10 mL pH= 2.78
at .02M and 1mL pH=3.32
a) Determine the [H3O+]eq from the pH values for each solution.
b) Construct ICE tables and use the initial molarity of [HC2H3O2] for each diluted solution, the equilibrium concentration for [H3O+]eq and the equation below to determine [C2H3O2–]eq and [HC2H3O2]eq.
HC2H3O2(aq) + H2O (l)--> H3O+(aq) + C2H3O2–(aq)
c. Write the Ka expression for the acid dissociation reaction and substitute the calculated equilibrium concentrations in to determine the Ka constant value.
d) Look up the accepted value of Ka for acetic acid in your textbook or other reference text. Calculate the % errror between this accepted value and your calculation.
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at .2M and 10 mL pH= 2.78 at .02M and 1mL pH=3.32 a) Determine the [H3O+]eq...
Acid Dissociation lab. Solution 1: 0.02M, pH3.20, solution 2: 0.2M pH3.67 Construct an ICE table and use the initial molarity of [HC2H3O2], the equilibrium concentration for [H3O+]eq and the equation; HC2H3O2-(aq)+H2O(l)=H3O+(aq)+C2H3O2-(aq) to determine HC2H3O2-]eq and [HC2H3O2]eq Can you please break it down step by step to I can use these examples to do other ICE problems and get the system down, thank you!
Important Info: Ka reaction: HC2H3O2(aq) + H2O(l) <-->C2H3O2 (aq) +H3O+(aq) HC2H3O(aq) +OH-(aq) -> C2H3O2-(aq) + H2O (l) A 1.0 L buffer solution is 0.280 M in HC2H302 (acetic acid) and 0.320 M in NaC2H302 (sodium acetate). Calculate the pH of the budder and he Ka for HC2H3O2 IS 1.8 X 10-5 Calculate the pH of te solution above after the addition of 0.0400 moles of solid NaOH. Assume no volume change upon the addition of base.
what is the hydronium ion concentration [H3O+] and the ph of a 0.5M acetic acid solution with Ka=1.8x10^-5? The equation for the dissociation of acetic acid is: CH3CO2H(aq) + H2O(l) = H3O + (aq) + CH3CO2-(aq)
1. Determine the volume, in mL, of 2.00 M
HC2H3O2 stock solution that would
need to be diluted to 50 mL in order to produce a solution that is
0.30 M in HC2H3O2.
2. Write the equilibrium constant expression, Ka, for
the ionization of acetic acid,
HC2H3O2.
3.In the rICE table for the dissociation of acetic acid, what is
the expression for acetic acid taken directly from the equilibrium
line of your ICE table? Do not use the x is...
Determine the molarity of an acetic acid solution, [HC2H3O2], with a pH of 2.87 (Ka for HC2H3O2 = 1.7 x 10-5). What is the initial concentration of an NH3 solution, [NH3], with a pH of 11.57 and a percent ionization value of 5%?
You are instructed to create 180. mL of a 0.06 M Acetate buffer with a pH of 4.0. You have Acetic acid and the sodium salt NaC2H3O2, available. (Enter all numerical answers to three significant figures.) HC2H3O2(s) + H2O(l) equilibrium reaction H3O+(aq) + C2H3O2−(aq) Ka1 = 1.7 ✕ 10−5 What is the molarity needed for the base component of the buffer? How many moles of base are needed for the buffer? How many grams of base are needed for the...
The following information and reaction equation will be used for the folliowing problems. A 120 mL solution of 0.110M acetic acid is going to be titrated with a series of additions of 0.100 M NaOH. (Acetic actic has a Ka= 1.8x10-5) HC2H3O2 (aq) + H2O (aq) <-----> H3O+ (aq) + C2H3O2- (aq) HC2H3O2 (aq) + NaOH (aq) <-----> H2O (aq) + NaC2H3O2 (aq) a) What is the pH of the above system after 66 mL of the 0.100 M NaOH...
I'm having trouble with the calculations for my chemistry lab: Determination of an Acid Dissociation Constant, Ka Half-Neutralization Molarity of acetic acid: 2 M Molarity of NaOH: 1 M Volume of acetic acid: 25 mL NaOH Solution Final Buret Reading: 6.23 mL Initial Buret Reading: 0.01 mL Volume added: 6.22 mL Total Volume of Solution: 250 mL pH: 3.85 Caculating Ka Initial number of moles HAn: ? OH-: ? Number of moles at equilibrium: HAn: ? An-: ? Equilibrium concentrations,...
If the a mixture of methylamine (CH3NH2, p Kb = 3.32) in water produces a pH of 11.8, what is the initial molar concentration of methylamine in the solution? O A 2.6x10-4M O B 7.7x10-2M O c 3.4x10-5M O D 8.3x10-2M Triethylamine (C2H5)3N) has a pKb = 3.25. What is the value of the equilibrium constant K for this reaction: (C2H5)3NH(aq) + H2O(1) = (C2H5)3N(aq) + H3O+(aq)? O A 1.77x10-11 O B 5.62x10-4 o C More information is needed to...
The Ka value for acetic acid, CH3COOH(aq), is 1.8×10−5. Calculate the pH of a 2.40 M acetic acid solution. pH= Calculate the pH of the resulting solution when 2.50 mL of the 2.40 M acetic acid is diluted to make a 250.0 mL solution. pH=