'Given 1.760 L of 0.384 M acid and 0.834 L of 0.724 base, find the final pH. Both the acid and base are strong.'
First I multiplied the volume by the molarity to find the moles of each.
Then I came up with this equation:
HX + YOH --> H2O + YX
(Y isn't yttrium; it's just some base Y)
And I made an ICE chart:
HX_____YOH_____YX
0.676___0.604___0
-0.604__-0.604__+0.604
0.072_____0_____0.604
I figured that since there was excess acid, that would determine the pH, so I took -log(0.072) and got a pH of about 1.143.
I'd just like to know if I'm doing this right, as there are give more problems like this and I don't want to do all of them only to find I'm messing up somewhere.
Thanks!
The thing in my mind is are the base, acid monoprotic?
help. I dont see a way.
Unfortunately I don't know. The problem only calls them a strong base and a strong acid, so I assumed they were monoprotic.
There is no way without knowing. If you know both are monoprotic it works ok. If both are diprotic it works OK, too. except that there are two H or OH in the excess reagent.
To determine the final pH of the solution, you need to consider the reaction between the acid (HX) and base (YOH). Since both the acid and base are strong, they will react completely, resulting in the formation of water (H2O) and a salt (YX).
First, let's calculate the moles of acid and base by multiplying their respective volumes by their molarities:
Moles of acid = volume of acid * molarity of acid = 1.760 L * 0.384 M = 0.676 moles
Moles of base = volume of base * molarity of base = 0.834 L * 0.724 M = 0.604 moles
Now, let's consider the reaction between the acid and base. The acid donates a proton (H+) to the base, resulting in the formation of water and the salt:
HX + YOH --> H2O + YX
Using the moles obtained, we can construct an ICE (Initial, Change, Equilibrium) table:
HX + YOH --> H2O + YX
Initial: 0.676 0.604 0
Change: -0.604 -0.604 +0.604
Equilibrium: 0.072 0 0.604
From the table, we can see that there is excess acid remaining after the reaction. The moles of acid remaining at equilibrium is 0.072.
To determine the final pH, you need to find the concentration of H+ ions in the solution. Since the volume of the solution is not given, we assume it remains constant. Therefore, the concentration of H+ ions is:
[H+] = (moles of acid remaining) / (volume of solution)
In this case, we don't have the volume, but we can assume it to be the sum of the volumes of acid and base:
Volume of solution = volume of acid + volume of base = 1.760 L + 0.834 L = 2.594 L
[H+] = (0.072 moles) / (2.594 L) = 0.028 M
Finally, to find the pH, take the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log [H+] = -log(0.028) ≈ 1.555
Therefore, the final pH of the solution is approximately 1.555.
It's important to note that the final pH depends on various factors, such as the strength and concentration of the acid and base, as well as the volumes used. Be careful in determining the volumes and molarities and double-check your calculations to ensure accuracy.