Question

question 3

Just as pH is the negative logarithm of [H3O+],

p*K*a is the negative logarithm of *K*a,

p*K*a*=*−log*K*a

The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions:

pH=p*K*a+log[base][acid]

Notice that the pH of a buffer has a value close to the
p*K*a of the acid, differing only by the logarithm of the
concentration ratio [base]/[acid]. The Henderson-Hasselbalch
equation in terms of pOH and p*K*b is similar.

pOH=p*K*b+log[acid][base]

Part A

Acetic acid has a *K*a of 1.8×10−5. Three acetic
acid/acetate buffer solutions, A, B, and C, were made using varying
concentrations:

a [acetic acid] ten times greater than [acetate],

b [acetate] ten times greater than [acetic acid], and

c [acetate]=[acetic acid].

Match each buffer to the expected pH.

Drag the appropriate items to their respective bins.

pH = 3.74

pH = 4.74

pH = 5.74

Part B

How many grams of dry NH4Cl need to be added to 1.70 L of a
0.200 *M* solution of ammonia, NH3, to prepare a buffer
solution that has a pH of 8.66? *K*b for ammonia is
1.8×10−5.

Express your answer with the appropriate units.

Answer #1

**part A**

**pH=pKa+log[base][acid]**

**pka of aceticacid = -logka**

**
= -log(1.8*10^-5)**

**
= 4.74**

**a. pH = 4.74+log(1/10)**

**pH = 3.74**

**b. pH = 4.74+log(10/1)**

** pH = 5.74**

**c. pH = 4.74+log(1/1)**

**pH = pka = 4.74**

**part B**

**pOH=pKb+log[acid][base]**

**[acid] = NH4+ = x
[base] = NH3 = 1.7*0.2 = 0.34 mol**

**pkb = -logkb**

** = -log(1.8*10^-5)**

** = 4.74**

**14-8.66 = 4.74 +log(x/0.34)**

**x = 1.3536**

**no of mol of NH4Cl = x = 1.3536 mol**

**mass of NH4Cl = 53.5*1.3536**

**
= 72.42 grams**

Acetic acid has a Ka of 1.8×10−5. Three acetic
acid/acetate buffer solutions, A, B, and C, were made using varying
concentrations:
[acetic acid] ten times greater than [acetate],
[acetate] ten times greater than [acetic acid], and
[acetate]=[acetic acid].
Match each buffer to the expected pH.
pH = 3.74 ; pH = 4.74 ; pH = 5.74
Part B: How many grams of dry NH4Cl need to be added to 2.30 L
of a 0.600 M solution of ammonia, NH3, to...

(You may use the Henderson-Hasselbalch equation to perform the
following calculations, but you do not have to.) The Ka of acetic
acid is 1.8 10–5. Review your calculations with your TA or
instructor before preparing the buffer solutions in lab. (Note:
solid sodium acetate comes as a hydrate (NaC2H3O2·3H2O), and thus
has a molar mass of 135.08 g/mol.) Buffer A: Calculate the mass of
sodium acetate that must be added to make 100.0 mL of an acetic
acid/acetate buffer...

The Henderson-Hasselbalch equation relates the pH of a buffer
solution to the pKa of its conjugate acid and the ratio of
the concentrations of the conjugate base and acid. The equation is
important in laboratory work that makes use of buffered solutions,
in industrial processes where pH needs to be controlled, and in
medicine, where understanding the Henderson-Hasselbalch equation is
critical for the control of blood pH.
Part A
As a technician in a large pharmaceutical research firm, you need...

The Henderson-Hasselbalch equation relates the pH of a buffer
solution to the pKa of its conjugate acid and the ratio of
the concentrations of the conjugate base and acid. The equation is
important in laboratory work that makes use of buffered solutions,
in industrial processes where pH needs to be controlled, and in
medicine, where understanding the Henderson-Hasselbalch equation is
critical for the control of blood pH.
Part A.) As a technician in a large
pharmaceutical research firm, you need...

You want to make 100 mL of 0.20 M Acetic Acid buffer with pH=4.0
You are given a stock solution of 1.0 M acetic acid and a bottle of
sodium acetate salt (MW= 82 g/mol). The formula for the
dissociation of acetic acid is shown here (CH3COOH <-->
CH3COO- + H+) The henderson hasselbach equation is : pH=pKa +log
[A-]/[HA].
What is the ratio of [A-]/[HA] when your buffer pH is 4.0?
Determine the concentration of weak acid and conjugate...

1.Write the equilibrium constant expression (Ka) for the generic
weak acid HA.
HA(aq)⇌H+(aq) + A−(aq)
2.Write the Henderson-Hasselbalch equation.
3.Given the Henderson-Hasselbalch equation, under what
conditions does the pH= pKa?
4.Sketch a pH versus volume of base curve (a titration curve)
for the titration of a weak acid with a strong base. On this sketch
indicate
the equivalence point and the point at which the conditions
described in #3 are met.
5.When using a buret, do your results depend on...

1. Calculate the hydrogen ion concentration, [H+ ],
for the two weak acids (pH=-log[H+ ], or [H+
]=antilog (-pH). If you have difficulty finding or using the
antilog function on your calculator, simply use this:
[H+ ]=10-pH .
2. Calculate the hydroxide ion concentration, [OH- ],
for the weak base using this formula: pOH=14-pH, then
[OH- ]=antilog (-pOH) or
[OH-]=10-pOH.
3. Calculate the molar concentrations of the vinegar as well as
ammonia. Both are industry standard 5.00% by mass solutions...

The Henderson-Hasselbalch Equation is: pH = pKa + log (
[A-]/[HA] ). Here is phosphoric acid at pH 0. It is polyprotic with
pKas of 2.12, 7.21, and 12.67 At what pHs would the average charge
on the phosphate species be -0.5, -1.0, and -1.5?

Using the Henderson-Hasselbalch equation:
1. What is the calculated pH of a solution containing 5mL of
4.5% of Acetic acid (C2H4O2) and
5mL of 0.500M of Sodium
acetate(C2H3NaO2)?
2. What is the calculated pH of a solution containing 5mL of
4.5% of Acetic acid (C2H4O2) and
1mL of 0.500M of Sodium acetate
(C2H3NaO2)?
Assuming that the solutions of acetic acid and sodium acetate
are equimolar.
HELP!!! Please show me how to do these calculations.

Use the Henderson Hasselbalch equation to calculate the pH of a
buffer solution prepared by dissolving 4.200 g of sodium acetate
and adding 8.5 mL of 6.0 M acetic acid in enough water to prepare
100.0 mL of
solution.

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