A solution was prepared from 10.0 mLof 0.100Mcacodylic acid and 10.0mLof 0.0800MNaOH. To this mixture was added 1.00mL of 1.27×10-6M morphine. Calling morphine B, calculate the fraction of morphine present in the form BH+.

Short Answer

Expert verified

The fraction of morphine Bin the form BH+ is 0.96

Step by step solution

01

Definition of Henderson-Hasselbalch equation.

  • To determine the pH = pKa + log([A]/[HA]) of a buffer, utilize the Henderson-Hasselbalch equation.
  • The equilibrium concentrations of the conjugate acid-base pair utilized to form the buffer solution are [HA] and [A] in this equation.
  • The pH of the solution is equal to the acid's pKa when [HA] = [A] . Jay came up with the idea.
02

Calculate the moles 10.0mL of 0.100M cacodylic acid and 10.0mL of 0.800MNaOH

A solution prepared from 10mL of 0.1M cacodylic acid (HA) and 10mL of 0.08MNaOH To this mixture was added 1mL of 1.27×10-6Mmorphine (B) . We need to calculate the fraction of morphine present in the form BH+.

First calculate the moles of:

a) 10mL of 0.1M cacodylic acid (HA)

nHA=c×VnHA=0.1M×10mLnHA=1mmol

b) 10mL of 0.08MNaOH

nOH-=c×VnOH-=0.08M×10mLnOH-=0.8mmol

03

Determine the value of  pKa

Consider the reaction

HA+OH-A-+H2O

Initial mmol:

HA=1mmolOH-=0.8mmolA-=0.8mmol

Final mmol:

HA=0.2mmolA-=0.8mmol

Determine the value of pKafor cacodylic acid (HA) by the following:

pKa=-logKapKa=-log6.4×10-7pka=6.19

Next calculate the pH of cacodylic acid (HA) using the Henderson-Hasselbach equation

pH=pKa+logA-/HApH=6.19+log0.8/0.2pH=6.79

04

Determine the value of  pKa of morphine A & B

Then calculate the Kafor morphine (B) :

Ka=-Kw/KbKa=10-14/1.6×10-6Ka=6.25×10-9

which gives us the pKafor morphine (B) :

pKa=-logKapKa=-log6.25×10-9pKa=8.20

05

Calculate the fraction of morphine B

Determine the quotient by the Henderson-Hasselbach equation:

pH=pKa+logB/BH+6.79=8.20+logB/BH+-1.41=logB/BH+10-1.41=B/BH+0.039=B/BH+

Next write the fraction for morphine (B) in the form of BH+:

α=BH+B+BH+α=BH+0.039BH++BH+α=0.96

Therefore, the fraction of morphine B is 0.96

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