A 0.5-L unbuffered solution needs the pH adjusted from 6 to 3. How many microliters of a 2 molar HCl solution need to be added to adjust the pH?
Report your answer to four decimal places.
For this question:
We know that: pH = -log([H+])
From this: [H+] = 10-pH
Initial [H+] = 10-6 M
Final [H+] = 10-3 M
Change in [H+] = 10-3 - 10-6 = 9.99 * 10-5 M
Moles of H+ = [H+] * Volume of solution = 9.99 * 10-5 * 0.5 = 5 * 10-5 moles
Molarity of HCl * Volume of HCl = moles of H+
2 * V = 5 * 10-5
Solving, V = 2.5 * 10-5 = 25 microliters
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