Calculate [H3O] in the following solutions.
a)4.0×10−2M HCl and 8.0×10−2M HOCl
b)0.105M NaNO2 and 5.00×10−2M HNO2
c)5.15x10-2M HCl and 7.62x10-2M NaC2H3O2
I would do the following:
a) This is a weak acid an a strong acid. I would calculate the Cl^- from HOCl with a common ion of H^+ from HCl, and recognize that (Cl^-) = (H^+) from HOCl. Then determine the H^+ from HCl, and add the two H^+ together.
b)This is a common ion problem or you can use the Henderson-Hasselbalch equation.
c)HCl + NaC2H3O2. It appears there is an excess of H^+ which will react with NaC2H3O2 to form HC2H3O2. That gives you a buffer of C2H3O2 and HC2H3O2 and you use the H-H equation to solve for pH.
Post your work if you get stuck.
a) Well, this question is certainly "acid"ic, but I'll try to keep my "pHunny" side up! Now, to calculate [H3O+] in the first solution, you need to find the concentration of H3O+ ions. In this case, because HCl is a strong acid, it completely dissociates into H3O+ and Cl-. So, the concentration of H3O+ ions will be the same as the concentration of HCl, which is 4.0×10^−2 M. And for HOCl, it partially dissociates, so you would need to use the acid dissociation constant (Ka) to find the concentration of H3O+ ions. But my "chlorine"telligence is telling me that you forgot to provide the Ka value for HOCl! So, sadly, I can't calculate [H3O+] for the second part of this solution. Keep in mind that chemistry can be quite "pHrustrating" sometimes!
b) Now let's move on to the next solution! For NaNO2, it's the salt of a weak acid, so it will not contribute any H3O+ ions to the solution. As for HNO2, it's a weak acid, which means it will partially dissociate. To determine the [H3O+], you would need to use the acid dissociation constant. But hey, "nitrous"-ly, you forgot to provide the Ka value for HNO2! Without that information, I'm "stumped" once again. So, no luck calculating [H3O+] for this solution. Don't worry, there's always a "mole-cule" of fun to have in chemistry, even when we can't calculate something!
c) Ah, the last solution! For HCl, it's a strong acid, so it completely dissociates into H3O+ and Cl-. Therefore, the concentration of H3O+ ions will be the same as the concentration of HCl, which is 5.15x10^-2 M. Now, for NaC2H3O2, it's a salt of a weak acid, so it will not contribute any H3O+ ions. So, no need to calculate anything here! Phew, finally an answer! "Sodium" to say, that chemistry is a balancing act between seriousness and laughter!
To calculate the [H3O+] (concentration of hydronium ions) in each solution, we need to consider the dissociation of acids and bases.
a) In the first solution, HCl is a strong acid that fully dissociates in water. The dissociation equation is:
HCl -> H+ + Cl-
Since HCl is a strong acid, it will completely dissociate into H+ ions and Cl- ions. Therefore, the concentration of [H3O+] is equal to the concentration of the acid, which is 4.0×10^(-2) M.
b) In the second solution, we have NaNO2 and HNO2. NaNO2 is a salt, so it does not contribute to the [H3O+] concentration. However, HNO2 is a weak acid that partially dissociates in water. The dissociation equation is:
HNO2 -> H+ + NO2-
To calculate [H3O+], we need to consider the concentration of HNO2 and the equilibrium constant (Ka) for its dissociation. From the information given, we do not have the Ka value. Without the Ka value, we cannot determine the exact [H3O+] concentration.
c) In the third solution, we have HCl and NaC2H3O2. Similar to the previous example, HCl is a strong acid that fully dissociates into H+ and Cl- ions. Therefore, the concentration of [H3O+] is equal to the concentration of the acid, which is 5.15x10^(-2) M.
On the other hand, NaC2H3O2 is a salt of a weak acid. C2H3O2- (acetate) is the conjugate base of acetic acid (CH3COOH). When a weak acid salt is dissolved in water, the acetate ions (C2H3O2-) can react with water to form acetic acid (CH3COOH) and hydroxide ions (OH-):
C2H3O2- + H2O -> CH3COOH + OH-
We need the Ka value for the acetic acid (CH3COOH) to calculate the [H3O+] concentration accurately. Without the Ka value, we cannot determine the exact [H3O+] concentration.
In summary, without the Ka values, we cannot calculate the exact [H3O+] concentrations for solutions b) and c).