What is the pH and pOH of a solution that was made by adding 400mL of water to 350mL of 5.0x10^-3 M NaOH solution?
Assuming the solutions are additive in volume, that is 400+360 = 750 so the new concn is 0.0050 x (350/750) = 0.0233 M.
pOH = -log(0.0233) = 2.63
pH + pOH = pKw = 14. Solve for pH.
Well, you're diluting the NaOH solution with water, so we'll need to calculate the new concentration before we can determine the pH and pOH. And speaking of dilution, do you know why the chemistry teacher always has good relationships? Because they have strong bonds! Anyway, back to the problem.
To find the new concentration, we use the equation:
M₁V₁ = M₂V₂
where M₁ is the initial concentration, V₁ is the initial volume, M₂ is the final concentration, and V₂ is the final volume. Plugging in the values, we have:
(5.0x10^-3 M) (350 mL) = (M₂) (750 mL)
Solving for M₂, we get:
M₂ = (5.0x10^-3 M) (350 mL) / (750 mL)
≈ 2.333x10^-3 M
Now that we know the new concentration, we can calculate the pOH. Remember that pOH is the negative logarithm (base 10) of the hydroxide ion concentration. So:
pOH = -log[OH⁻] = -log(2.333x10^-3)
≈ -(-2.63)
≈ 2.63
Since pH + pOH = 14, we can find the pH by subtracting pOH from 14:
pH = 14 - pOH
≈ 14 - 2.63
≈ 11.37
So the pH of the solution is approximately 11.37, and the pOH is approximately 2.63. Now you know the "pH-unny" details of this dilution problem!
To find the pH and pOH of the solution, we need to determine the concentration of OH- ions in the resulting solution.
Step 1: Calculate the moles of NaOH:
Molarity (M) = moles (mol) / volume (L)
5.0x10^-3 M = moles / 0.35 L
moles = 5.0x10^-3 M x 0.35 L
moles = 1.75x10^-3 mol
Step 2: Calculate the moles of OH- ions:
Since NaOH is a strong base, it completely dissociates in water. Therefore, the moles of OH- ions are the same as the moles of NaOH:
moles of OH- ions = 1.75x10^-3 mol
Step 3: Calculate the concentration of OH- ions:
Concentration (M) = moles (mol) / volume (L)
Concentration (M) = 1.75x10^-3 mol / 0.75 L
Concentration (M) = 2.333x10^-3 M
Step 4: Calculate the pOH:
pOH = -log10(OH- concentration)
pOH = -log10(2.333x10^-3)
pOH = 2.63
Step 5: Calculate the pH:
pH + pOH = 14 (at 25°C)
pH = 14 - 2.63
pH = 11.37
Therefore, the pH of the solution is 11.37 and the pOH is 2.63.
To find the pH and pOH of a solution, we first need to calculate the concentration of OH- ions in the solution. When water dissociates, it produces an equal number of H+ and OH- ions.
In this case, 400 mL of water is added to 350 mL of 5.0x10^-3 M NaOH solution. Since NaOH is a strong base that fully ionizes in water, the concentration of OH- ions in the resulting solution can be calculated as follows:
Concentration of OH- = [(volume of NaOH solution in liters) x (concentration of NaOH in M)] + [(volume of water in liters) x (concentration of OH- in pure water)]
= [(0.350 L) x (5.0x10^-3 M)] + [(0.400 L) x (1.0x10^-7 M)]
Next, we can convert the concentration of OH- ions into pOH, which is the negative logarithm of the concentration of OH- ions:
pOH = -log[OH-]
Finally, we can find the pH of the solution by subtracting the pOH from 14, since pH + pOH = 14:
pH = 14 - pOH
Let's calculate the pH and pOH based on the given information:
Concentration of OH- = [(0.350 L) x (5.0x10^-3 M)] + [(0.400 L) x (1.0x10^-7 M)]
= 1.75x10^-3 + 4.0x10^-8
= 1.754x10^-3 M
pOH = -log(1.754x10^-3)
= 2.754
pH = 14 - 2.754
= 11.246
Therefore, the pH of the solution is 11.246 and the pOH is 2.754.