If 25.5 grams of NaOH is dissolved in 750 ml of water, find:
[H+]=
[OH-]=
pH=
pOH=
moles NaOH = grams/molar mass
(NaOH) = moles/L soln.
pOH = -log(OH^-)
pH + pOH = pKw = 14.
To find the values of [H+], [OH-], pH, and pOH, we need to use the concept of moles and the equation that relates concentration, volume, and number of moles.
1. Start by calculating the number of moles of NaOH:
- Moles = Mass / Molar mass
- The molar mass of NaOH = 22.99 g/mol (Na) + 15.999 g/mol (O) + 1.008 g/mol (H) = 39.997 g/mol
- Moles = 25.5 g / 39.997 g/mol = 0.6375 mol
2. Calculate the concentration of NaOH (in moles per liter):
- Concentration (Molarity) = Moles / Volume
- Volume = 750 ml = 0.75 L
- Concentration = 0.6375 mol / 0.75 L = 0.85 M
3. Since NaOH is a strong base, it fully dissociates in water:
- NaOH → Na+ + OH-
- Therefore, the concentration of OH- is the same as the concentration of NaOH, which is 0.85 M.
4. The concentration of H+ and OH- in water at room temperature is related by the equation:
- [H+] × [OH-] = 1.0 x 10^-14 (at 25°C)
5. Calculate the concentration of H+:
- [H+] = 1.0 x 10^-14 / [OH-] = 1.0 x 10^-14 / 0.85 M = 1.176 x 10^-15 M
6. Calculate the pH:
- pH = -log[H+]
- pH = -log(1.176 x 10^-15) = 14.93
7. Calculate the pOH:
- pOH = -log[OH-]
- pOH = -log(0.85) ≈ 0.07
Therefore, the values are:
[H+] ≈ 1.176 x 10^-15 M
[OH-] = 0.85 M
pH ≈ 14.93
pOH ≈ 0.07