Question

1. Show the calculations required for preparing the buffers in this exercise. The pKa values for H3PO4 are 2.15, 6.82, 12.32. Note that you’ll be using the pKa value of 6.82. Why?

Method 1 a. Calculate the grams of NaH2PO4*H2O needed to prepare 100.0 mL of a 100mM solution. b. Calculate the amount of 1.00M NaOH you expect to add to adjust the pH of your NaH2PO4 solution to 7.0.

Method 2 c. Using the Henderson-Hasselbalch equation, calculate the g of NaH2PO4*H2O and g of Na2HPO4 needed to prepare 100.0 mL of 100.0mM sodium phosphate buffer, pH 7.0.

2. Describe how you would prepare the following solutions: a. 100.0 mL of 1-butanol:formic acid:water (100:35:25) b. 100.0 mL of 0.025M potassium hydrogen phosphate, 0.025 M disodium hydrogen phosphate, 0.100M NaCl

Help Please!

Answer #1

1) a. 100 mM = 100 mmol NaH2PO4.H2O/1 L solution

i.e. 10 mmol NaH2PO4.H2O is present in 100 mL solution

The molar mass of NaH2PO4.H2O = 138 g mol^{-1}

i.e. The mass of NaH2PO4.H2O = 10*10^{-3} mol * 138 g
mol^{-1} = 1.38 g

b and c. According to Henderson-Hasselbulch equation:

pH = pKa2 + (n_{NaOH}/n_{NaH2PO4.H2O})

Here, you need to adjust pH to 7.

i.e. 7 = 6.82 + Log(10 mmol/x mL*1 mmol/mL)

i.e. Log(10/x) = 7 - 6.82

i.e. 10/x = 10^{0.18}

i.e. x = 10/1.5

i.e. The volume of 1 M NaOH needed to adjust pH of the solution to 7 is 6.6 mL

Note: Here, you need to use pKa = 6.82 because the equilibrium
exists between NaH_{2}PO4 and Na2HPO4

i.e. H_{2}PO_{4}^{-}
HPO_{4}^{2-}

You need to prepare a buffer for biochemistry lab. The required
solution is 0.5Msodium phosphate, pH 7.0. Use the
Henderson-Hasselbalch equation to calculate the number of moles and
grams of moobasic sodium phosphate ( NaH2PO4) and dibasic sodium
phosphate ( Na2HPO4) necessary to make 1 liter of solution. The pKa
for this buffer is 7.21

Phosphoric acid is a triprotic acid with the following pKa
values: pka1= 2.148 pka2= 7.198 pka3= 12.375
You wish to prepare 1.000 L of a 0.0300 M phosphate buffer at pH
7.680. To do this, you choose to mix the two salt forms involved in
the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric
flask and add water to the mark. What mass of each salt will you
add to the mixture?
What other combination of phosphoric acid...

0.10 M Phospate buffer at pH 7.0 total 100.0 ml. 30mL of 0.20M
NaH2PO4, 20 mL of 0.20M Na2HPO4 and 50.0 mL of water were used to
prepare the buffer.
The average pKa is 6.76
Using the average pKa value calculated in 3, calculate the pH of
the pH 7.0 buffer solution that is expected after the addition of
1.0 mL of 1.0M HCl.

Show calculation for the assigned buffer. Recall: [acid]
+ [base] = 0.20 M. Calculate the individual concentrations
needed (mass or volume) for HA and A- based on your assigned buffer
and pH. The condition [acid] + [base] = 0.20 M
must be satisfied. The final volume of solution is
100.0 mL.
Assigned Buffer: sodium dihydrogen phosphate and disodium
hydrogen phosphate
pH = 7.69
Supplemental Data (I believe Ka is 6.2
x10^-8)
Acid
Species
Ka
Molar Mass g/mol
Compound
Phosphoric
density...

A) Need to prepare 100 ml of a 0.10 M phosphate buffer with a pH
of 6.2. Calculate the amounts of 1.0 M NaH2PO4 and 1.0 M K2HPO4
solutions which will give the required pH of 6.2.Use the
Henderson-Hasselbalch equation ( pH = pKa + log (base/acid)
B) Need to prepare 100 ml of a solution containing 0.025 M
NaHCO3 and 0.12 M NaHCO3 and 0.12 M NaCl From stock solutions of
0.05 M NaHCO3 and 0.24 M NaCl from...

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...

I have already posted this questions once... but some doofus
responded with "additional information needed". But I posted all
the information that the question gives, word for word. So i'm
going to give it another shot. Here is a similar question to Part
A.
(http://www.chegg.com/homework-help/questions-and-answers/technician-large-pharmaceutical-research-firm-need-produce-450ml-100m-potassium-phosphate--q1392759)
± Buffers in Medicine
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...

Write a procedure so a lab mate could prepare 100 mL of buffer
solution of that pH from two solids. Your procedure should clearly
indentify which glassware and chemicals you will use to prepare the
solution. Assume that you will use the sodium salt of the conjugate
base to prepare the solution. Our desired pH is 7.5 and the weak
acid I chose is Sodium hydrogen sulfite with pKa of 7.21.
Use Henderson-Hasselbalch equation to determine molequantities
of the weak...

1. What volume of 0.100 M HCl is required to neutralize 0.15 g of
sodium acetate?
2.
What mass of potassium hydrogen phthalate, KHP is needed to
neutralize 40.00 mL of 0.100 M NaOH?
3.
What is the pH of the mixture if 10.00mL of 0.100 M NaOH is added
to a 20.00 mL, 0.10M ascorbic acid solution?
4.
What mass of a weak acid with a molar mass of 100.0 g/mol is
necessary to neutralize 25.0 mL of 0.100M...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 6 minutes ago

asked 42 minutes ago

asked 48 minutes ago

asked 56 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago