The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.
Answer:
pH = 3.513
Explanation:
Hello there!
In this case, since this titration is carried out via the following neutralization reaction:
We can see the 1:1 mole ratio of the acid to the base and also to the resulting acidic salt as it comes from the strong HCl and the weak hydroxylamine. Thus, we first compute the required volume of HCl as shown below:
Now, we can see that the moles of acid, base and acidic salt are all:
And therefore the concentration of the salt at the equivalence point is:
![[HONH_3^+Cl^-]=\frac{0.0044mol}{0.022L+0.0293L} =0.0858M](https://tex.z-dn.net/?f=%5BHONH_3%5E%2BCl%5E-%5D%3D%5Cfrac%7B0.0044mol%7D%7B0.022L%2B0.0293L%7D%20%3D0.0858M)
Next, for the calculation of the pH, we need to write the ionization of the weak part of the salt as it is able to form some hydroxylamine as it is the weak base:
Whereas the equilibrium expression is:
![Ka=\frac{[H_3O^+][HONH_2]}{[HONH_3^+]}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BHONH_2%5D%7D%7B%5BHONH_3%5E%2B%5D%7D)
Whereas Ka is computed by considering Kw and Kb of hydroxylamine:
So we can write:
And neglect the x on bottom to obtain:
And since x=[H3O+] we obtain the following pH:
Regards!
Hmm, I am not quite sure but if I am probably your best bet so I'd go with, either B or D.
The answer is D. Neutral net charge
Answer:-3463 kJ and -3452kJ
Explanation:
ΔU is the change in internal energy of a system and its formula is;
ΔU = q + w
Where q represents heat transferred into or out of the system. Its value is positive when heat is transfer into the system and negative when heat is produced by the system.
W represents the work done on or by the system. Its value is positive when work is done on the system and negative when it is done by the system.
For the system in this question, we see that it produces heat which means heat is transferred out of the system, therefore the value of q is negative, it can also be seen that work is done by the system which means that w is also negative.
Therefore,
ΔU = -q-w
ΔU = -3452 kJ – 11kJ
= - 3463kJ
ΔH is the change in the enthalpy of a system and its formuls is;
ΔH = ΔU + Δ(PV)
By product rule Δ(PV) becomes ΔPV + PΔV
At constant pressure ΔP = 0. Therefore,
ΔH = -q-w + PΔV
w is equals to PΔV, So:
ΔH = -q
ΔH = -3452kJ
