Answer :
Final answer:
The final pH of the buffer solution, after adding a strong base, is 4.75 (only slightly different from the initial pH).
Explanation:
A buffer solution is prepared by combining a weak acid (HX) with its conjugate base (NaX). In this case, 37.9 mL of 2.62 M HX and 51.5 mL of 5.07 M NaX are utilized. The pH of the buffer solution, prior to the addition of a strong base (NaOH), can be calculated using the equation provided.
After adding 3.76 mL of 1.12 M NaOH, the final pH of the mixture is found to be 4.75 (only slightly different from the initial pH).
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Final answer:
The final pH of the buffer solution, after addition of NaOH, is calculated using the Henderson-Hasselbalch equation, considering the moles of weak acid, conjugate base, and the added strong base. The final calculated pH is 6.14.
Explanation:
The question is asking about the pH of a buffer solution when a strong base (NaOH) is added. First, calculate the moles of HX (weak acid), X- (conjugate base), and OH- (strong base) present in the solution. This is done by multiplying the volume (in liters) by the molarity of each:
HX: 37.9 mL x 2.62 M/1000 = 0.099 moles
X-: 51.5 mL x 5.07 M/1000 = 0.261 moles
OH-: 3.76 mL x 1.12 M/1000 = 0.00421 moles
Next, since OH- is a strong base, it will react completely with HX to form X- and H2O until either OH- or HX is depleted. Here, HX is in excess (0.099 - 0.00421 moles = 0.095 moles remain), and the number of moles of X- will also increase (0.261 + 0.00421 moles = 0.265 moles).
You can then use the Henderson-Hasselbalch equation, which states that pH = pKa + log([salt]/[acid]), where pKa = -log(Ka).
pKa = -log(5.8 × 10-6) = 5.24, and
pH = 5.24 + log(0.265/0.095) = 5.24 + 0.9 = 6.14.
Therefore, the final pH of the mixture after all three components are added is 6.14.
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