pOH when there are 1.6 ×10^-7 moles of H+in a 500ml of solution
To find the pOH of the solution, we first need to find the concentration of hydroxide ions (OH-) in the solution. Since water undergoes autoionization to form equal concentrations of H+ and OH- ions, the concentration of OH- ions can be found using the concentration of H+ ions.
Given that there are 1.6 ×10^-7 moles of H+ ions in a 500 ml solution, we can calculate the concentration of H+ ions as follows:
n = 1.6 × 10^-7 moles
V = 500 ml = 0.5 L
[ H+ ] = n / V
[ H+ ] = (1.6 × 10^-7) / 0.5
[ H+ ] = 3.2 × 10^-7 M
Since the concentration of H+ and OH- ions are equal in water, the concentration of OH- ions is also 3.2 × 10^-7 M.
Now, we can find the pOH of the solution using the formula:
pOH = -log[OH-]
pOH = -log(3.2 × 10^-7)
pOH ≈ -log(3.2) - log(10^-7)
pOH ≈ -0.5052 + 7
pOH ≈ 6.4948
Therefore, the pOH of the solution is approximately 6.49.