You are instructed to create 300. mL of a 0.34 M phosphate buffer with a pH of 7
ID: 946199 • Letter: Y
Question
You are instructed to create 300. mL of a 0.34 M phosphate buffer with a pH of 7.3. NaH2PO4 is the acid component of the buffer. Na2HPO4 is the base component of your buffer.
What is the molarity needed for the acid component of the buffer?
What is the molarity needed for the base component of the buffer?
How many moles of acid are needed for the buffer?
How many moles of base are needed for the buffer?
How many grams of acid are needed for the buffer?
How many grams of base are needed for the buffer?
Explanation / Answer
The ratio of acid component : base component in the phosphate buffer with pH = 7.3 is calculated using
Handerson's- Hasselbalch's equation
pH = pKa + log [NaH2PO4] / [ Na2HPO4]
substituting the given values we get,
log { [ Na H2 PO4 ] / [Na2HPO4} } = pH - pKa*...........*pKa of NaH2PO4 = 7.2
.........................................................= 7.3 - 7.2 = 0.1
& [ NaH2PO4] / [ Na2HPO4]..........= 1.2589
hence the molarity of acid component = 1.2589 M
................................base component = 1.000 M
But we are instructed to prepare 300ml. of o.34 M buffer solution-
This can be found out by determining the volumes of acid and base components needed separately using relation
M1V1 = M2V2
Thus volume of acid component .........= ( 300 x 0.34 ) / 1.2589
.............................................................= 81.0 ml
hence moles of acid component needed = (81.0 x 1.2589) / 1000
................................................................ = 0.102 mols
and weight of acid component ................ = (0.102 x 120)
................................................................. = 12.24 gms.
Similarly, mols of base component is calculated as = 0.102 mols
...................weight of base component .....................= 14.48 gms.
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