How many moles of H3O+ or OH-
must you add per liter of HA solution to adjust its pH from 9.75 to 9.01? Assume a negligible volume change.
……….x 10………Moles of ……………………..h3O+ or
OH-
I got 4.6001 x 10-5 M of h3o+...but it was wrong
Well, it seems like you ended up in mole-trouble with your calculation. Don't worry, I'm here to clown around and help you out!
To adjust the pH from 9.75 to 9.01, we're looking at a decrease in acidity. Remember, a decrease in pH means an increase in the concentration of H3O+ ions.
Let's break it down step by step:
1. Calculate the difference in H3O+ concentration:
10^(-pH) = [H3O+]
[H3O+] initial = 10^(-9.75)
[H3O+] final = 10^(-9.01)
2. Determine the change in concentration:
Δ[H3O+] = [H3O+] final - [H3O+] initial
3. Convert the change in moles:
Moles H3O+ = Δ[H3O+] x volume of solution in liters
Remember, since the volume change is considered negligible, we can assume it remains constant.
So, keep going and calculate the change in moles of H3O+ ions, and hopefully, you'll get the right answer this time! And don't forget to stay hydrated, even if we're talking about acidic solutions!
To calculate the number of moles of H3O+ or OH- required to adjust the pH from 9.75 to 9.01, you can use the formula:
[H+] = 10^(-pH)
First, calculate the concentration of H+ ions (or H3O+ ions) at pH 9.75:
[H+] = 10^(-9.75)
[H+] = 1.78 x 10^(-10) M
Similarly, calculate the concentration of H+ ions (or H3O+ ions) at pH 9.01:
[H+] = 10^(-9.01)
[H+] = 7.93 x 10^(-10) M
Since the initial concentration of H+ ions is 1.78 x 10^(-10) M and you need to increase it to 7.93 x 10^(-10) M, you can calculate the difference:
Δ[H+] = (7.93 x 10^(-10) M) - (1.78 x 10^(-10) M)
Δ[H+] = 6.15 x 10^(-10) M
Since H3O+ and OH- ions have a 1:1 relationship in water, the concentration of OH- ions required will also be 6.15 x 10^(-10) M.
Therefore, to adjust the pH from 9.75 to 9.01, you need to add 6.15 x 10^(-10) moles of H3O+ or OH- ions per liter of HA solution.
To calculate the moles of H3O+ or OH- needed to adjust the pH of a solution, you can use the equation:
pH = -log[H3O+]
First, calculate the concentration of H3O+ initially in the solution with a pH of 9.75:
[H3O+] = 10^(-pH)
[H3O+] = 10^(-9.75)
Next, calculate the concentration of H3O+ you want to achieve, based on the desired pH of 9.01:
[H3O+] = 10^(-pH)
[H3O+] = 10^(-9.01)
Now, subtract the initial concentration from the desired concentration to find the change in concentration:
Change in [H3O+] = [H3O+]_desired - [H3O+]_initial
Finally, multiply the change in concentration by the volume of the solution to find the moles of H3O+ or OH- needed per liter:
Moles = Change in [H3O+] x Volume
Since H3O+ and OH- are always in a 1:1 ratio in water, the moles of OH- will also be the same as the moles of H3O+.
Let's calculate it:
[H3O+]_initial = 10^(-9.75) = 1.77828 x 10^(-10) M
[H3O+]_desired = 10^(-9.01) = 7.91823 x 10^(-10) M
Change in [H3O+] = (7.91823 x 10^(-10)) - (1.77828 x 10^(-10)) = 6.13995 x 10^(-10) M
Moles = (6.13995 x 10^(-10) M) x 1 L = 6.13995 x 10^(-10) moles
So, you need approximately 6.13995 x 10^(-10) moles of H3O+ or OH- per liter of HA solution to adjust its pH from 9.75 to 9.01.
Check this out.
pH = 9.75 gives (H^+) = 1.77E-10 mols/L
pH = 9.01 gives (H^+) = 9.77E-10 mols/L.
Difference is 8.00E-10 so you want to increase the (H3O^+) by 8.00E-10 mols/L. Do you see anything wrong with that?
figure the H+ concentration of each
1) H+ =10^-9.75=1.7782794e-10
2) H=10^-9.01=9.7723722e-10
so the difference between these two concentrations is 7.9940928e-10