Here’s the math for your answer, which is 3.3 L HCl

Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

The mixture would contain
if
undergoes no hydrolysis; the solution is of volume
after the mixing. The two species would thus be of concentration
and
, respectively.
Construct a RICE table for the hydrolysis of
under a basic aqueous environment (with a negligible hydronium concentration.)

The question supplied the <em>acid</em> dissociation constant
for acetic acid
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant
for its conjugate base,
. The following relationship relates the two quantities:

... where the water self-ionization constant
under standard conditions. Thus
. By the definition of
:
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)


![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
Answer : The concentration of
is, 
Explanation :
When we assume this reaction is driven to completion because of the large excess of one ion then we are assuming limiting reagent is
and
is excess reagent.
First we have to calculate the moles of KSCN.


Moles of KSCN = Moles of
= Moles of
= 
Now we have to calculate the concentration of ![[Fe(SCN)]^{2+}](https://tex.z-dn.net/?f=%5BFe%28SCN%29%5D%5E%7B2%2B%7D)
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{\text{Moles of }[Fe(SCN)]^{2+}}{\text{Volume of solution}}](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%7D%7B%5Ctext%7BVolume%20of%20solution%7D%7D)
Total volume of solution = (6.00 + 5.00 + 14.00) = 25.00 mL = 0.025 L
![\text{Concentration of }[Fe(SCN)]^{2+}=\frac{1.08\times 10^{-5}mol}{0.025L}=4.32\times 10^{-4}M](https://tex.z-dn.net/?f=%5Ctext%7BConcentration%20of%20%7D%5BFe%28SCN%29%5D%5E%7B2%2B%7D%3D%5Cfrac%7B1.08%5Ctimes%2010%5E%7B-5%7Dmol%7D%7B0.025L%7D%3D4.32%5Ctimes%2010%5E%7B-4%7DM)
Thus, the concentration of
is, 
Weathering breaks down rock and other material
The mole fraction of helium is 0.33.
The mole fraction can be calculated with the aid of dividing the number of moles of 1 factor of an answer by means of the overall number of moles of all the additives of a solution. it's far noted that the sum of the mole fraction of all the components in the solution should be same to one.
Mole fraction is a unit of attention. within the answer, the relative amount of solute and solvents are measured by the mole fraction and it is represented. The mole fraction is the wide variety of moles of a specific thing within the solution divided by way of the entire wide variety of moles.
Mole fraction represents the quantity of molecules of a selected thing in a combination divided by using the total variety of moles within the given mixture. it is a manner of expressing the concentration of a solution. consequently, the sum of mole fraction of all of the additives is always equal to one.
Mole of neon = 1
mole of helium = 2
mole of argon = 3
Total mole = 1 + 2 + 3 = 6
Mole fraction = mole of helium / total mole
= 2 / 6 = 0.33
Learn more about mole fraction here:-brainly.com/question/1601411
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