Question

In: Chemistry

Concept questions: 1) Acetic acid is a weak acid. Why is a 2M solution of acetic...

Concept questions: 1) Acetic acid is a weak acid. Why is a 2M solution of acetic acid less hazardous than a 2M solution of hydrochloric acid? 2) What species must be present in a buffer? Write a chemical reaction that explains how the buffer reacts with acid and then write a chemical reaction that shows how the buffer reacts with base. Wxplain how a buffer works.

Solutions

Expert Solution

Question 1

Acetic acid is a weak acid given that is a substance whose dissociation grade is low in other words is a substance forming a low quantity of ions when is dissolve in water. While hydrochloric acid is a strong acid since is a substance capable of dissociating completely when is dissolved in water.

Due to this property a 2M solution of  hydrochloric acid is more hazardous compared to a solution of the same concentration of acetic acid.  Hydrochloric acid is more corrosive given that generates a higher quantity of ions which can cause more damage.

Question 2

In a buffer solution two species must be present a weak base or a weak acid and its respective salt. For example a solution formed by CH3COOH and CH3COOK it's a buffer solution.

A buffer solution can regulate the changes in pH when an acid is added due to the presence of a conjugate base. For example if we have a solution constitued by CH3COOH (weak acid) and CH3COO- (conjugate base), the acetic acid will disociate as follows:

If a little quantity of a strong acid is added, this substance provides H+ ions to the solution which react with the conjugate base, in this case the CH3COOH- ions produced by the former reaction as showed bellow:

In the case a little quantity of a strong base is added, this substance provides OH- ions which react with the acid, and hence the pH remains almost constant given that all of this ions are neutralized by the acid as indicated bellow:


Related Solutions

1) Answer the following the questions regarding an acetic acid buffer solution. a) A solution was...
1) Answer the following the questions regarding an acetic acid buffer solution. a) A solution was prepared by dissolving 0.024 moles of acetic acid (HOAc; pKa =4.8 ) in water to give 1 liter of solution. What is the pH? b)Then 0.009 moles of sodium hydroxide (NaOH) was added in to the solution. What is the pH of the solution? (In this problem, you may ignore changes in volume due to the addition of NaOH) c)Additionally, 0.015 moles of NaOH...
A solution contains 1.5M sodium acetate. The pKa of acetic acid (a weak acid) is 4.76....
A solution contains 1.5M sodium acetate. The pKa of acetic acid (a weak acid) is 4.76. Calculate the fraction and percentage of the molecules existing in the unionized state.
pH of acetic acid solutions. Remember, acetic acid is a weak acid, with a Ka =...
pH of acetic acid solutions. Remember, acetic acid is a weak acid, with a Ka = 1.8 x 10−5. Obtain 10.0 mL of 0.10 M acetic acid and place in a clean dry 50.0 mL beaker. Predict the value of the pH. Measure the pH with the pH meter. Record the value. Take 1.00 mL of the 0.10 M CH3COOH(aq) in the previous step, and put it in another clean beaker. Add 9.00 mL of deionized water and stir. What...
If you started with an acetic acid solution of 2.0 mL of 0.84M Acetic acid and...
If you started with an acetic acid solution of 2.0 mL of 0.84M Acetic acid and shifted it with 0.2mL of 0.01M HCl (4 drops), what is the new pH?. Initial pH Final pH Ka of acetic acid = 1.8 x 10-50.3 Start by finding the equilibrium concentrations of the acetic acid solution. Calculate the Ph based on this initial concentration (initial Ph).   Calculate the moles of everything (compounds in the equilibrium AND the acid). Shift the equilibrium with the...
1.) Acetic acid is a weak monoprotic acid with Ka=1.8*10^-5. In an acid base titration, 100...
1.) Acetic acid is a weak monoprotic acid with Ka=1.8*10^-5. In an acid base titration, 100 mL of 0.100M acetic acid is titrated with 0.100M NaOH. What is the pH of the solution: a.)Before any NaOH is added b.)Before addition of 15.0mL of 0.100M NaOH c.)At the half-equivalence point d.)After addition of total of 65.0mL of 0.100M NaOH e.)At equivalence point f.)After addition of a total of 125.0mL of 0.100M NaOH
Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x...
Acetic acid is a monoprotic weak acid with a pKa of 4.74. (Ka = 1.8 x 10-5) (a) What is the pH of 5 mL of a 5.0% solution? (b) What is the pH of the solution if you now add 45 ml of water to solution (a)? How will the equivalence point volumes differ if you titrate the two solutions ?
1.) Determine the initial concentration of a solution of the weak acid HClO2 , if it...
1.) Determine the initial concentration of a solution of the weak acid HClO2 , if it its pH is 1.20 and Ka = 1.10
Calculate the percent ionization of 1.50 M aqueous acetic acid solution. For acetic acid, Ka =...
Calculate the percent ionization of 1.50 M aqueous acetic acid solution. For acetic acid, Ka = 1.8 × 10−5 . (a) 2.71% (b) 3.55% (c) 1.78% (d) 0.35% (e) None of the above
If pKa of weak acids are the following: pKa acetic acid = 4.75 pKa propionic acid...
If pKa of weak acids are the following: pKa acetic acid = 4.75 pKa propionic acid = 4.87 pKa acitric acid = 5.40 pKa of benzoic acid = 4.19 What is the order of acids with increasing acidity Determine the pH of the following A. 100 mL of benzoic acid solution 0.03M is mixed with 200 ml water B. 50 mL of sodium hydroxide solution 0.05 M is mived with 200 ml water
Codeine (C18H12NO3) is a weak base, (pKb= 5.79). A solution of 5*10^-2M solution has a pH...
Codeine (C18H12NO3) is a weak base, (pKb= 5.79). A solution of 5*10^-2M solution has a pH of 10. Calculate the [C18H22NO3], [C18H21NO3] and [OH-] at equilibruim.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT