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11 Dec 2019

Please help!**** for each of the indicated concentration strong or weak acids determine concentrations of acid [HA], hydronium ion H3O and pH of the solution (ka,Kb values on table)

15mL of 0.500 M HCL mixed with water to make 2.00 L of solution

and

10mL of 0.50 M acetic acid mixed with water to make 25.0 ml solution .

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13 Dec 2019

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One beaker contains 10.0 mL of acetic acid/sodium acetate buffer at maximum buffer capacity (equal concentrations of acetic acid and sodium acetate), and another contains 10.0 mL of pure water. Calculate the hydronium ion concentration and the pH after the addition of 0.25 mL of 0.10 M HCl to each one. What accounts for the difference in the hydronium ion concentrations? Explain this based on equilibrium concepts; in other words, saying that âone solution is a bufferâ is not sufficient. H3O+(aq) + C2H3O2-(aq) â HC2H3O2(aq) +H2O Ka of acetic acid = 1.8 x 10-5 Acetic acid/sodium acetate buffer: Acetic acid = 0.1M, Sodium acetate = 0.1M

For the dissociation reaction of a weak acid in water,

HA(aq)+H2O(l)âH3O+(aq)+Aâ(aq)

the equilibrium constant is the acid-dissociation constant, Ka, and takes the form

Ka=[H3O+][Aâ][HA]

Weak bases accept a proton from water to give the conjugate acid and OHâ ions:

B(aq)+H2O(l)âBH+(aq)+OHâ(aq)

The equilibrium constant Kb is called the base-dissociation constant and can be found by the formula

Kb=[BH+][OHâ][B]

When solving equilibrium-based expression, it is often helpful to keep track of changing concentrations is through what is often called an I.C.E table, where I. stands for Initial Concentration, C. stands for Change, and E. stands for Equilibrium Concentration. To create such a table, write the reaction across the top creating the columns, and the rows I.C.E on the left-hand side.

Initial (M)Change (M)Equilibrium (M)A+ BâAB

Part A

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolved 2.00 g of aspirin in 0.600 L of water and measured the pH. What was the Ka value calculated by the student if the pH of the solution was 2.60?

Express your answer numerically using two significant figures.

Hints

Ka =

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You used the initial concentration of aspirin in the Ka expression. Instead, you need to use the equilibrium concentration. Consider that the amount of aspirin that reacted is equal to the amount of H₃O⁺ produced. How much is left over?

Part B

A 0.100 M solution of ethylamine (C2H5NH2) has a pH of 11.87. Calculate the Kb for ethylamine.

Express your answer numerically using two significant figures.

Hints

Kb =

Take approximately 20 mL of 0.5 M of HC2H3O2 (acetic acid) solution in a clean, small beaker. Prepare 25 mL of three different acetic acid solutions with concentrations in the range between 0.05 and 0.2 M from the stock 0.5 M acetic acid solution using appropriate dilution techniques. (Use glass pipettes and volumetric flasks). Determine the pH of each solution using the pH meter as described below.

Concentrations used are .05,0.1,and 0.15

Data-

Write the chemical equation for the dissociation of acetic acid.

Prepare and complete an ICE table for each solution of acetic acid tested:

	[HC2H3O2] (M)	[C2H3O2-] (M)	[H3O+] (M)
Initial Concentration		0	0
Concentration Change
Equilbrium Concentration

Determine the Ka and pKa values for each solution of acetic acid tested. Show your work

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