Use the References to access important values If needed for this question Design
ID: 706033 • Letter: U
Question
Use the References to access important values If needed for this question Design a buffer that has a pH of 6.92 using one of the weak base/conjugate acid systems shown below. Kb Conjugate Acid K Weak Base CH NH2 4.2x10 CHNH 2.4x10 CHsO,N 59x0 iO,NH 1.7x10 CHAN 1.5x10° C,HsNH' 6.7x 106 pKa 10.62 7.77 5.17 base, to How many grams of the chloride salt of the conjugate acid must be combined with how many grams of the weak produce 1.00 L of a buffer that is 1.00 M in the weak base? grams chloride salt of conjugate acid grams weak baseExplanation / Answer
The pH of the solution is 6.92. A buffer must be chosen so that the pKa of the acid component in the buffer satisfies the relation (pH – 1) < pKa < (pH + 1). Since, only C6H15O3NH+ has pKa = 7.77 and falls in the range (pH – 1) < pKa < (pH + 1), hence, C6H15O3NH+ (as the chloride salt) and C6H15O3N are chosen for the buffer.
Use the Henderson-Hasslebach equation to calculate the ratio of C6H15O3N and C6H15O3NH+ in the buffer.
pH = pKa + log [C6H15O3N]/[C6H15O3NH+]
====> 6.92 = 7.77 + log [C6H15O3N]/[C6H15O3NH+]
====> -0.85 = log [C6H15O3N]/[C6H15O3NH+]
====> [C6H15O3N]/[C6H15O3NH+] = antilog (-0.85) = 0.1412
====> [C6H15O3N] = 0.1412*[C6H15O3NH+] ……(1)
It is given that the concentration of the weak base in the buffer is 1.00 M, i.e, the [C6H15O3N] = 1.00 M. Therefore,
1.00 M = 0.1412*[C6H15O3NH+]
====> [C6H15O3NH+] = (1.00 M)/(0.1412) = 7.0821 M.
We have 1.00 L of the buffer solution; hence,
moles of C6H15O3NH+ = (volume of buffer)*(molarity of C6H15O3NH+) = (1.00 L)*(7.0821 M) = 7.0821 mole.
moles of C6H15O3N = (volume of buffer)*(molarity of C6H15O3N) = (1.00 L)*(1.00 M) = 1.00 mole.
Molar mass of C6H15O3N = (6*12.011 + 15*1.008 + 3*15.999 + 1*14.007) g/mol = 149.190 g/mol.
Molar mass of C6H15O3NH+ (as the chloride salt), i.e, C6H15O3NHCl = (6*12.011 + 16*1.008 + 3*15.999 + 1*14.007 + 1*35.453) g/mol = 185.651 g/mol.
Grams chloride salt of conjugate acid = (moles of conjugate acid)*(molar mass of conjugate acid) = (7.0821 mole)*(185.651 g/mol) = 1314.7989 g ? 1314.8 g (ans).
Grams weak base = (moles of weak base)*(molar mass of weak base) = (1.00 mole)*(149.190 g/mol) = 149.190 g ? 149.2 g (ans).
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