Calculate [H3O+] in the following solutions.
1) 0.100 M NaNO2 and 5.00×10−2 M HNO2
2) 5.20×10−2 M HCl and 7.62×10−2 M NaC2H3O2
please help thx
I'm not going to work four problems this long for you. I'll do #1 and get you started on #2.
NaNO2 is determined by hydrolysis of the NO2^- and HNO2 is a weak acid.
................NO2^- + HOH ==>HNO2 + OH^-
I...............0.1...............................0...........0
C...............-x................................x............x
E.............0.1-x..............................x............x
Kb for NO2^- = (Kw/Ka for HNO2) = (x)(x)/(0.1-x)
Substitute and solve for x = (OH^-) and convert to (H3O^+)
For HNO2, that is a weak acid and
...............HNO2 ==> H^+ + NO2^-
I..............0.05............0.........0
C.............-x................x.........x
E..........0.05-x.............x.........x
Write the Ka expession for HNO2, plug in the E line and solve for x = H^+ = H3O^+
The object, I think, for having you do these two (in #1) is to show you that the salt is basic but the acid is acidic.
#2. HCl is a strong acid; therefore, (H^+) = (HCl) The acidity of the NaC2H3O2 (sodium acetate) is determined by the hydrolysis of the acaetate ion. That's done exactly the same way as the nitrite ion in #1.
Post your work if you get stuck.
I've calculate part 1)
................NO2^- + HOH ==>HNO2 + OH^-
I...............0.1...............................0...........0
C...............-x................................x............x
E.............0.1-x..............................x............x
Kb for NO2^- = (Kw/Ka for HNO2) = (x)(x)/(0.1-x)
4.5x10^-4 = (x)(x)/(0.1-x)
x = 6.75x10^-3 M
x = [OH-]
-log [OH-]
pH= 14- (-log [OH-])
pH =11.8
For HNO2, that is a weak acid and
...............HNO2 ==> H^+ + NO2^-
I..............0.05............0.........0
C.............-x................x.........x
E..........0.05-x.............x.........x
4.5x10^-4 = x^2 / 0.05-x
x = 4.52x10^-3
pH = 2.3
so which is the [H3O+] ? FOR QUESTION #1
To calculate the [H3O+] concentration in the given solutions, we need to consider the dissociation of the relevant acids and bases.
1) 0.100 M NaNO2 and 5.00×10−2 M HNO2:
The relevant reaction for HNO2 is:
HNO2 + H2O ⇌ H3O+ + NO2-
Since NaNO2 is a salt, it dissociates completely in water and does not contribute to the acid-base reaction.
To calculate the [H3O+] concentration, we need to determine the equilibrium concentration of HNO2. To do that, we need to first solve for the Ka value of HNO2:
Ka = [H3O+][NO2-] / [HNO2]
The Ka value for HNO2 is 4.0 x 10^-4.
[HNO2] is given as 5.00×10−2 M.
Since the initial concentration of HNO2 is the same as its equilibrium concentration, we can assume the change in concentration of H3O+ and NO2- is x. Therefore, the change in concentration of HNO2 is -x.
At equilibrium, [HNO2] = 5.00×10−2 - x
[NO2-] = 0 + x
[H3O+] = 0 + x
Substituting the values into the Ka expression:
4.0 x 10^-4 = (x)(x) / (5.00×10−2 - x)
Since x is much smaller than 5.00×10−2, we can make the approximation (5.00×10−2 - x) ≈ 5.00×10−2.
4.0 x 10^-4 ≈ (x^2) / (5.00×10−2)
Rearranging the equation:
x^2 ≈ 4.0 x 10^-4 * 5.00×10−2
x^2 ≈ 2.0 x 10^-7
Taking the square root of both sides, we get:
x ≈ 1.41 x 10^-4
Thus, [H3O+] ≈ 1.41 x 10^-4 M.
2) 5.20×10−2 M HCl and 7.62×10−2 M NaC2H3O2:
The relevant reaction for HCl is:
HCl + H2O ⇌ H3O+ + Cl-
The relevant reaction for NaC2H3O2 is:
CH3COONa + H2O ⇌ CH3COOH + OH-
Since HCl is a strong acid, it will dissociate completely, and we can assume [H3O+] ≈ [HCl]. Thus, [H3O+] ≈ 5.20×10−2 M.
There is also the formation of OH- by the base C2H3O2-, but since the concentration of OH- is not given, we cannot determine its concentration at this point.
Please note that the concentration of Na+ and Cl- ions will not contribute to the [H3O+] concentration.
Therefore, the [H3O+] concentration in the second solution is approximately 5.20×10−2 M.
To calculate the concentration of [H3O+] in the given solutions, we need to understand the concept of acid-base equilibrium and use the expression for the ionization constant of water.
1) Solution with NaNO2 and HNO2:
- First, let's write down the balanced chemical equation for the dissociation of HNO2 in water:
HNO2 + H2O ⇌ H3O+ + NO2-
- We know that HNO2 is a weak acid, so it partially ionizes in water.
- The concentration of H3O+ can be determined by calculating the concentration of HNO2 that has ionized, using the acid dissociation constant (Ka) for HNO2.
- The Ka expression for HNO2 is: Ka = [H3O+][NO2-]/[HNO2]
Now, let's calculate the [H3O+] for the given solution:
- Given: [HNO2] = 5.00×10−2 M (concentration of HNO2 in the solution).
- Assume x is the concentration of HNO2 that ionizes, which will also be the concentration of [H3O+] formed.
- At equilibrium: [HNO2] - x = [H3O+]
- Since x is assumed to be small, we can neglect it in the [HNO2] - x expression.
- Ka = [H3O+][NO2-]/[HNO2] ⟹ Ka = [H3O+]²/[HNO2]
- Rearrange the equation to solve for [H3O+]: [H3O+]² = Ka × [HNO2]
- Substitute the given values: [H3O+]² = Ka × [HNO2] = Ka × 5.00×10−2 M
- Now, find the value of Ka from the literature. For HNO2, the Ka value is 4.5×10−4.
Using the given values and the calculated Ka, we can solve for [H3O+]:
[H3O+]² = (4.5×10−4)(5.00×10−2) M
[H3O+]² ≈ 2.25×10−5 M
[H3O+] ≈ √(2.25×10−5) M
[H3O+] ≈ 4.74×10−3 M
Therefore, the concentration of [H3O+] in the solution is approximately 4.74×10−3 M.
2) Solution with HCl and NaC2H3O2:
- First, let's write down the balanced chemical equation for the dissociation of HCl in water:
HCl + H2O → H3O+ + Cl-
- HCl is a strong acid that completely ionizes in water.
- So, the concentration of [H3O+] in the solution is equal to the concentration of HCl.
Therefore, the concentration of [H3O+] in the solution is 5.20×10−2 M (same as the concentration of HCl).
Note: In the second solution, H3O+ concentration is directly provided because HCl is a strong acid that completely ionizes in water.