Question

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1. You are asked to prepare 10.00 mL of a buffer solution consisting of 0.15 M weak acid (HA) and 0.15 M conjugate base (A™). The target pH of the buffer solution is 4.50. The K of HA is 8.2 x 10-5. What volume of A- is required to make the buffer? You must show your work to receive credit. (3 pts.) Cypka s logo 1. 1 1 . .1

Answer 1

From Henderson-Hasselbalch equation:

$pH=pK_a+log\frac{[conjugate\, base]}{[acid]}$

So, here,

$pH=pK_a+log\frac{[A^-]}{[HA]}$

Now, say we have to mix y mL of A^-, thus, volume of HA must be (10-y) mL.

Now, y mL 0.15 M A^- is equivalent to 0.15y mL.M = 0.15y mL. mol/L = 0.15y (mL/L).mol = 0.15y x 10^-3 mol = 0.15y mmol

Similarly, (10 - y) mL 0.15 M HA is equivalent to 0.15(10-y) mmol HA.

Now, pK_a = - log K_a = - log (8.2 x 10^-5) = 4.09

Since, concentration is proportional to no of moles, thus in HH equation we can substitute the concentration terms with mmoles.

$4.50=4.09+log\frac{0.15y}{0.15(10-y)}$

$log\frac{y}{10-y}=0.41$

$Or,\frac{y}{10-y}=10^{0.41}=2.5704$

$Or,\frac{y}{10-y+y}=10^{0.41}=\frac{2.5704}{1+2.5704}=0.72$

$Or,\frac{y}{10}=0.72$

$Or,y=7.2$

So, we have to take 7.2 mL of A^-.