Calculate the pH of a 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCL. (0.1 M HCl and NaOH)
In total, you need to show 10 calculations.
One example each for 0.1M HCl/NaOH added to 25 mL H2O.
The diluted HCl solution is 0.1M x (1 mL/26 mL) = (HCl)
pH = -log(HCl)
The diluted NaOH solution is 0.1M x (1 mL/26 mL) = (NaOH)
pOH = -log(NaOH).
Then pH + pOH = pKw = 14.00
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To calculate the pH of a sample after the addition of NaOH and HCl, we need to understand how these compounds affect the pH.
NaOH is a strong base, while HCl is a strong acid. When added to water, NaOH dissociates to produce hydroxide ions (OH-) and HCl dissociates to produce hydrogen ions (H+). The concentration of H+ and OH- ions in a solution determines its pH.
Let's start the calculations:
1. For 1 mL of NaOH:
The moles of NaOH = volume (in L) × molarity = 0.001 L × 0.1 M = 0.0001 mol.
The moles of OH- produced = moles of NaOH = 0.0001 mol.
The concentration of OH- ions = moles / volume = 0.0001 mol / 0.026 L = 0.0038 M.
The pOH = -log(OH-) = -log(0.0038) ≈ 2.42.
The pH = 14 - pOH = 14 - 2.42 ≈ 11.58.
2. For 1 mL of HCl:
The moles of HCl = volume (in L) × molarity = 0.001 L × 0.1 M = 0.0001 mol.
The moles of H+ produced = moles of HCl = 0.0001 mol.
The concentration of H+ ions = moles / volume = 0.0001 mol / 0.026 L = 0.0038 M.
The pH = -log(H+) = -log(0.0038) ≈ 2.42.
Repeat the above steps for each volume of NaOH and HCl:
3. For 2 mL of NaOH:
The concentration of OH- ions = moles / volume = 0.0002 mol / 0.026 L ≈ 0.0077 M.
The pOH = -log(OH-) = -log(0.0077) ≈ 2.11.
The pH = 14 - pOH = 14 - 2.11 ≈ 11.89.
4. For 2 mL of HCl:
The concentration of H+ ions = moles / volume = 0.0002 mol / 0.026 L ≈ 0.0077 M.
The pH = -log(H+) = -log(0.0077) ≈ 2.11.
Repeat the above steps for 3 mL, 4 mL, and 5 mL of NaOH and HCl to get the pH for each case.
5. For 3 mL of NaOH:
The concentration of OH- ions = 0.0115 M.
The pOH = 2.94.
The pH = 14 - pOH ≈ 11.06.
6. For 3 mL of HCl:
The concentration of H+ ions = 0.0115 M.
The pH = 2.94.
7. For 4 mL of NaOH:
The concentration of OH- ions = 0.0154 M.
The pOH = 2.81.
The pH = 14 - pOH ≈ 11.19.
8. For 4 mL of HCl:
The concentration of H+ ions = 0.0154 M.
The pH = 2.81.
9. For 5 mL of NaOH:
The concentration of OH- ions = 0.0192 M.
The pOH = 2.72.
The pH = 14 - pOH ≈ 11.28.
10. For 5 mL of HCl:
The concentration of H+ ions = 0.0192 M.
The pH = 2.72.
These calculations will give you the pH of the 25 mL sample of distilled water after the addition of 1 mL, 2 mL, 3 mL, 4 mL, and 5 mL of NaOH and HCl.