In an experiment, a solution of pH 4.0 containing 100 mM K+ and 0.1 mM Ca++ is r

Question

In an experiment, a solution of pH 4.0 containing 100 mM K+ and 0.1 mM Ca++ is required. Choose the correct way to prepare this solution

from the following options.

a. Dissolve 3.73 g of KCl in 90 ml citrate buffer of pH 4.0, add 1 ml of 10 mM CaCl2, and make up the volume to 500 ml.

b. Dissolve 0.373 g of KCl in 40 ml distilled water.Add 0.5 ml of 10 mM CaCl2 to 5 ml distilled water. Mix both the solutions and adjust the pH to 4.0 with HCl. Make up the volume to 50 ml.

c. Dissolve 7.46 g of KCl in 50 ml distilled water. To this, add 10 ml of 10 mM CaCl2. Adjust the pH to 4.0 with HCl and make up the volume to 100 ml.

d. Dissolve 0.746 g of KCl in 100 ml of distilled water, add 1 ml of 10 mM CaCl2, and adjust the pH to 4.0 with HCl.

Please explain how did you come up with that answer along with the calculation. Thanks

Explanation / Answer

For the preparation of solution containing 100 mM OF K+ and 0.1 mM OF Ca2+ ions,

We can calculate the mili moles of Potassium chloride and mili moles of cacl2,

we know the formula for moles= wt. in. gms/mol.wt,

there fore to get wt. in gms, the formula will become,

wt. in gms= moles* molecular wt,

we know that ,milimoles/1000= moles, (milimoles=moles*1000),

KCl= mol wt= 74.55 , we required 100 mM,

SO, 100*74.55=7455/1000=7.455 GMS OF KCl,

for 0.1 mM OF cacl2 ,

cacl2= mol wt= 110.98,

0.1*110.98=11.098/1000=0.1109 gms of cacl2,

we can convert it to mmoles moles as,

for 100 mM k+ = 100/1000=0.1 moles, (0.1*74.55=7.455 gms KCl )

for 0.1 mM Ca2+ = 0.1/1000=0.0001 moles *110.98=0.01109 gms cacl2

from above information only option 'C' is correct.

FOR KCl= 7.46 gms and total volume make up to 100 ml,

molarity = moles/volume in L,

WE know the moles = 0.1 moles/0.10= 1 M, (100 ML=0.10 L) (7.46/74.55=0.100 MOLES*1000=100 mM)

MOLES= MOLARITY *VOLUME IN L,

MOLES= 1*0.10= 0.1 MOLES *1000=100 mM,

FOR CaCl2= 0.1 mMOLES= 0.0001 MOLES,

MOLARITY = MOLES/VOLUME IN L,

=0.0001/0.01 L, (VOLUME =10 ML= 0.01 L)

=0.01 Molar CaCl2,

moles= molarity *volume in L,

= 0.01*0.01=0.0001 MOLES*1000= 0.1 mM

SO THE ONLY OPTION 'C' is the correct answer.

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