SOLVE FOR part B AND C Please solve for B and C What is the pH of a buffer prepa
ID: 837814 • Letter: S
Question
SOLVE FOR part B AND C
Please solve for B and C
What is the pH of a buffer prepared by adding 0.708mol of the weak acid HA to 0.305mol of NaA in 2.00 L of solution? The dissociation constant Ka of HA is 5.66 times 10 -7. Express the pH numerically to three decimal places. What is the pH after 0.150mol of HC1 is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places. What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places. What is the pH after 0.150mol of HC1 is added to the buffer from Part A? Assume no volume change on the addition of the acid. Express the pH numerically to three decimal places. What is the pH after 0.195mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base. Express the pH numerically to three decimal places.Explanation / Answer
Let's start with part B then. We have a buffer containing 0.708 mol of HA and 0.305 mol of NaA (the conjugate base). When we add 0.150 mol of HCl, it will react with NaA to be neutralized. Specifically, it will convert 0.150 mol of NaA into the weak acid. So let's recalculate how many moles of HA and NaA we have.
0.305 mol NaA - 0.150 mol = 0.155 mol NaA remaining after reaction with HCl
0.708 mol + 0.150 mol = 0.858 mol HA now after reaction.
Now we are still in a buffer system, and all we need to do to find the pH is use the Henderson-Hasselbalch equation. Note the pKa = -log(Ka) = -log(5.66 x 10-7) = 6.25. Thus:
pH = pKa + log ( [A-] / [HA] )
pH = 6.25 + log ( 0.155 / .858 ) = 6.25 - 0.74 = 5.51
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For Part C, the same idea applies. Now we have 0.195 base to react with our weak acid and convert it into the conjugate base. Thus now we have:
0.708 mol HA - 0.195 mol = 0.513 mol HA remaining after addition of NaOH
0.305 mol A- + 0.195 mol = 0.5 mol A- after addition of NaOH
Plugging back into Henderson-Hasselbalch yields:
pH = 6.25 + log ( 0.5 / 0.513 ) = 6.25 - 0.01 = 6.24
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