What is the pH of the solution obtained when 125 mL of 0.614M NaOH is diluted to 16.0 L with water?
M NaOH = 0.614 x (125 mL/16,000 mL) = ??
Convert that to pOH and pH.
Oh, pH, the measure of how acidic or basic a solution is. Let me clown around with the numbers for a second. So, we start with 125 mL of 0.614 M NaOH, but when you dilute it with water, the amount of NaOH stays the same while the volume increases. It's like NaOH trying to do a disappearing act!
To calculate the new concentration, we can use the formula C₁V₁ = C₂V₂, where C₁ and V₁ are the initial concentration and volume, and C₂ and V₂ are the final concentration and volume.
Let's do some calculations in my imaginary clown lab. The initial concentration times the initial volume should be equal to the final concentration times the final volume. So, (0.614 M)(125 mL) = (C₂)(16,000 mL).
By solving for C₂, we find that the final concentration is approximately 0.00483 M. But wait, there's more!
Now, pH measures the negative logarithm of the concentration of hydrogen ions in a solution. Since NaOH is a base, it produces hydroxide (OH-) ions. So, to find the pH, we need to know the concentration of hydrogen ions.
Since NaOH is a strong base, it dissociates completely in water, and the concentration of hydroxide ions is equal to the concentration of NaOH. In this case, it's 0.00483 M.
Now, here comes the fun part. We use the inverse logarithm to calculate the hydrogen ion concentration, which is 1 divided by the hydroxide ion concentration. So, 1/0.00483 M gives us approximately 207.01. Drumroll, please!
Finally, we calculate the pH by taking the negative logarithm of the hydrogen ion concentration. That gives us pH = -log(207.01) ≈ -2.32.
So, my friend, the pH of this diluted solution is approximately -2.32. It's like a clown doing a science experiment – full of surprises!
To find the pH of the solution, we first need to calculate the concentration of the diluted NaOH solution.
Step 1: Calculate the final concentration of the diluted NaOH solution.
To do this, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of NaOH = 0.614 M
V1 = initial volume of NaOH = 125 mL = 0.125 L
C2 = final concentration of NaOH (Unknown)
V2 = final volume of the diluted solution = 16.0 L
Plugging in the values:
(0.614 M)(0.125 L) = C2(16.0 L)
C2 = (0.614 M)(0.125 L)/(16.0 L)
C2 = 0.00472 M
Step 2: Calculate the pOH of the solution.
Since NaOH is a strong base, we can use the equation pOH = -log(OH- concentration).
The concentration of OH- ions is equal to the concentration of NaOH in moles per liter.
pOH = -log(0.00472 M)
Step 3: Calculate the pH of the solution.
The pH can be found using the equation pH + pOH = 14.
pH + pOH = 14
pH + (-log(0.00472 M)) = 14
pH = 14 + log(0.00472 M)
Calculating the pH:
pH = 14 + log(0.00472 M)
pH ≈ 12.33
Therefore, the pH of the solution obtained after diluting 125 mL of 0.614 M NaOH to 16.0 L with water is approximately 12.33.
To find the pH of a solution obtained by diluting a strong base like NaOH, we need to use the equation for the hydrolysis of the base.
First, we need to determine the moles of NaOH in the original solution:
moles of NaOH = concentration (M) × volume (L)
= 0.614 M × 0.125 L
= 0.07675 mol
Since NaOH is a strong base, it fully dissociates in water to produce hydroxide ions (OH-). When we dilute this solution, the concentration of the hydroxide ions decreases, but the number of moles remains the same.
The number of moles of NaOH in the final solution also remains the same, so we can calculate the final concentration of hydroxide ions:
final concentration (M) = moles / volume (L)
= 0.07675 mol / 16.0 L
= 0.004797 M
Now, since we have the concentration of the hydroxide ions, we can find the pOH of the solution:
pOH = -log[OH-]
= -log(0.004797)
= 2.32
To find the pH, we subtract the pOH from 14 (since pH + pOH = 14):
pH = 14 - pOH
= 14 - 2.32
= 11.68
Therefore, the pH of the solution obtained when 125 mL of 0.614M NaOH is diluted to 16.0 L with water is approximately 11.68.