A solution is made by dissolving 26.9 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places.
I'm super confused about this one. I got the moles to be .6748 then divided that by .5 to get 1.3496 M. But then when I got the -log it was a negative answer. Why?
Oh, and this one: Calculate the pOH of 1.62 M HI.
I thought it was a simple matter of getting the -log and then subtracting from 14 to get 13.8 but that wasn't right.
Please help!
just an update: i got the first one but am uber confused about the second.
a.
mols NaOH = grams/molar mass = 26.9/40 = about 0.67 and that divided by 0.5 = about 1.35M
pH = -log 1.35
pH = -(0.13) = -0.13 approximately.
Remember 1.0M = zero pH.
b. And it is that simple. It's that negative pH again.
HI = 1.62M
(H^+) = 1.62M
pH = -log(1.62) = about -0.21
pH + pOH = 14
-0.21 + pOH = 14
pOH = 14-(-0.21) = 14.21.
To calculate the pH of the final solution, we need to understand the steps involved. Let's break it down step-by-step:
Step 1: Calculate the moles of NaOH:
Given mass of NaOH = 26.9 g
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol
Moles of NaOH = mass / molar mass = 26.9 g / 39.99 g/mol ≈ 0.673 moles
Step 2: Calculate the concentration of NaOH solution:
Given volume of water = 450 mL = 0.45 L
Concentration = moles / volume = 0.673 moles / 0.45 L ≈ 1.495 M
Step 3: Calculate the concentration of final solution:
Given final volume = 500 mL = 0.5 L
Concentration = 1.495 M (since no additional water is added in this step)
Step 4: Calculate the pOH:
To calculate the pOH, we need the concentration of the hydroxide ions (OH-). Since NaOH is a strong base, it completely dissociates in water, providing one hydroxide ion for each sodium ion. Therefore, the concentration of hydroxide ions (OH-) is equal to the concentration of NaOH.
pOH = -log [OH-] = -log 1.495 ≈ -log (1.5) = -0.18
Now, let's move on to the second question:
To calculate the pOH of 1.62 M HI, we use a similar approach:
Step 1: Calculate the pOH:
Given concentration of HI = 1.62 M
In this case, HI is a strong acid, so it completely dissociates in water, meaning the concentration of hydronium ions (H3O+) is equal to the concentration of HI.
pOH = -log [H3O+] = -log 1.62 ≈ -log (1.6) = -0.21
Remember that the pH and pOH are related by the equation: pH + pOH = 14. Therefore, to calculate the pH, you subtract the pOH from 14:
pH = 14 - pOH = 14 - (-0.21) ≈ 14.21
So, the pH of the final solution is approximately 14.21, and the pOH of 1.62 M HI is approximately 0.21.
To calculate the pH of a solution, you need to determine the concentration of hydrogen ions (H+). This concentration can be found by calculating the concentration of hydroxide ions (OH-) present in the solution.
In the first question, you correctly calculated the molarity (M) of the NaOH solution after it was diluted with water. However, to find the concentration of hydroxide ions (OH-) in the solution, you need to use the balanced chemical equation for the dissociation of NaOH in water:
NaOH → Na+ + OH-
From the balanced equation, you can see that the molar ratio of NaOH to OH- is 1:1. Therefore, the concentration of OH- ions is equal to the molarity of the NaOH solution.
In this case, the concentration of hydroxide ions (OH-) is 1.3496 M (as you correctly calculated).
Now, to calculate the concentration of hydrogen ions (H+), you can use the equation:
[H+] x [OH-] = 1.00 x 10^-14 (at 25°C)
Since the concentration of OH- ions is equal to the concentration of H+ ions, you can substitute this value in:
[H+]^2 = 1.00 x 10^-14
Taking the square root of both sides, you get:
[H+] = 1.00 x 10^-7 M
This means that the concentration of hydrogen ions (H+) in the solution is 1.00 x 10^-7 M, which is equal to the solution's pH.
Regarding the negative answer when calculating the -log of the concentration, it seems like you might have made a calculation error. The -log(1.00 x 10^-7) should be equal to 7, not a negative value. Make sure to double-check your calculations.
Now, moving on to the second question, to calculate the pOH of a solution, you need to determine the concentration of hydroxide ions (OH-). This concentration can be found by dividing the concentration of HI by the stoichiometric coefficient of the hydroxide ion in the dissociation reaction.
From the balanced equation for the dissociation of HI:
HI → H+ + I-
You can see that the stoichiometric coefficient of OH- is zero. Therefore, HI does not directly produce OH- ions in solution.
Instead, you need to use the equation:
pOH + pH = 14
Since you know the concentration of HI (1.62 M), you can use this equation to find the pH of the solution:
pH = 14 - pOH
pH = 14 - (-log(1.62))
Substituting the value of -log(1.62) into the equation, you should be able to correctly calculate the pH.
Remember to use parentheses when taking the -log of a value to ensure you are applying the logarithm function properly.