Question

The Henderson-Hasselbalch equation relates the pH of a buffer
solution to the p*K*a 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 to produce 350 mL of a
potassium dihydrogen phosphate buffer solution of pH = 6.99. The
p*K*a of H2PO4− is 7.21.

You have the following supplies: 2.00 L of 1.00 *M*
KH2PO4 stock solution, 1.50 L of 1.00 *M* K2HPO4 stock
solution, and a carboy of pure distilled H2O.

How much 1.00 *M* KH2PO4 will you need to make this
solution? (Assume additive volumes.)

**Volume of KH2PO4 needed = ?**

Physicians use the following
modified form of the Henderson-Hasselbalch equation to track
changes in blood pH:Carbon dioxide (CO2) and bicarbonate
(HCO3−) concentrations in the bloodstream are physiologically
controlled to keep blood pH constant at a normal value of 7.40.

pH=p*K*a+log ( [HCO3−] / (0.030)(*P*CO2) )

where [HCO3−] is given in millimoles/liter and the arterial
blood partial pressure of CO2 is given in mmHg. The p*K*a of
carbonic acid is 6.1. Hyperventilation causes a physiological state
in which the concentration of CO2 in the bloodstream drops. The
drop in the partial pressure of CO2 constricts arteries and reduces
blood flow to the brain, causing dizziness or even fainting.

**Part B.)** If the normal physiological
concentration of HCO3− is 24 m*M*, what is the pH of blood
if *P*CO2 drops to 27.0 mmHg ?

**pH= ?**

Answer #1

**A)**

**pH = pKa + log [ K2HPO4 / KH2PO4]**

**6.99 = 7.21 + log [ K2HPO4 / KH2PO4]**

**[ K2HPO4 / KH2PO4] = 0.6**

**let**

**volume of KH2P04 be y L**

**then**

**volume of K2HP04 = 0.35 - y**

**now**

**moles = molarity x volume**

**so**

**moles of KH2P04 = 1 x y = y**

**moles of K2HP04 = 1 x ( 0.35 - y) = 0.35 -
y**

**now**

**[ K2HPO4 / KH2PO4] = 0.6**

**(0.35 - y ) / y = 0.6**

**0.35 - y = 0.6y**

**0.35 = 1.6 y**

**y = 0.21875**

**so**

**volume of KH2P04 = 0.21875 L = 218.75 ml**

**so**

**218.75 ml of KH2P04 is required**

**B)**

**pH = pKa + log [HC03-] / 0.03 PCO2**

**given**

**pKa = 6.1**

**[HC03-] = 24 mM**

**PCO2 = 27 mm Hg**

**so**

**pH = 6.1 + log 24 / 0.03 x 27**

**pH = 7.57**

**so**

**the pH is 7.57**

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...

Use the Henderson - Hasselbalch equation to calculate the
proprotion in which two equal concentrations of H2PO4- (
acid) and HPO42- (conjugate base) should be mixed to
produce a phosphate buffer at a pH of 7. The pKa for
this acid-base pair is 6.86.

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.

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?

Use the Henderson- Hasselbalch equation to calculate the pH of
each solution.
a. a solution that is 0.145 M in propanoic acid and 0.115 M in
potassium propanoate
b. a solution that contains 0.785% C5H5N by mass and 0.985%
C5H5NHCI by mass
c. a solution that contains 15.0 g of HF and 25.0 g of NaF in
125 mL of solution

Use the Henderson-Hasselbalch equation to calculate the pH of
each solution:
Part A a solution that is 0.20 M in HCHO2 and 0.15 M in
NaCHO2
Part B a solution that is 0.16 M in NH3 and 0.22 M in NH4Cl

1.Use the Henderson-Hasselbalch equation to calculate the pH of
a 1 liter buffer made by mixing 0.135 M HClO and 0.155 M KClO
What will be the pH of the above buffer in question 1. if you
add 10.0 mL of 6.0 M
NaOH to the buffer ? What will be the pH if you add 10.0 mL of
5.0 M HCl instead ? ( can you show all steps and formulas)

Use the Henderson-Hasselbalch equation to calculate the Ph pf
each solution.
A) A solution that contains 1.00% C2H5NH2 by mass and 1.19%
C2H5NH3 Br by mass. Express your answer using two decimal places.
(pH=?)
B) A solution that is 13.0g HC2H3O2 and 11.5g of NaC2H3O2 in
150.0 mL of solution. Express your answet using two decimal places.
(pH=?)

question 3
Just as pH is the negative logarithm of [H3O+],
pKa is the negative logarithm of Ka,
pKa=−logKa
The Henderson-Hasselbalch equation is used to calculate the pH
of buffer solutions:
pH=pKa+log[base][acid]
Notice that the pH of a buffer has a value close to the
pKa of the acid, differing only by the logarithm of the
concentration ratio [base]/[acid]. The Henderson-Hasselbalch
equation in terms of pOH and pKb is similar.
pOH=pKb+log[acid][base]
Part A
Acetic acid has a Ka of 1.8×10−5....

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