I think it is A. ability to oxidize :)
HCl is a strong acid therefore complete ionisation takes place
HCl ---> H⁺ + Cl⁻
1 mol of HCl dissociates to form 1 mol of H⁺
NaOH is a strong base therefore complete ionisation takes place
NaOH ---> Na⁺ + OH⁻
1 mol of NaOH dissociates to form 1 mol of OH⁻ ions
equal number of moles of HCl and NaOH will dissociate to give out an equal number of moles of H⁺ and OH⁻ ions.
H⁺ + OH⁻ --> H₂O
equal amount of H⁺ and OH⁻ together forms H₂O
since all the H⁺ and OH⁻ have now reacted to form water molecules the solution now becomes neutral as there are no excess H⁺ ions to make the solution acidic and no excess OH⁻ ions to make the solution basic.
Therefore the solution will be neutral.
Answer : The mass of sodium sulfate needed is 5.7085 grams.
Explanation : Given,
Concentration of sodium ion = 0.148 mol/L
Volume of solution = 2.29 L
Molar mass of sodium sulfate = 142 g/mole
First we have to determine the moles of sodium ion.
Now we have to calculate the moles of sodium sulfate.
The balanced chemical reaction will be,
As, 2 moles of sodium ion produced from 1 moles of 
So, 0.08035 moles of sodium ion produced from 
moles of
Now we have to calculate the mass of sodium sulfate.
Therefore, the mass of sodium sulfate needed is 5.7085 grams.
If we abbreviate the formula for nicotine as Nic, then the equations for two different equilibria of Nic in water are
Nic + H2O ---> NicH+ + OH-
NicH+ + H2O ---> NicH2 2+ + OH-
We can write the Kb1 expression for the first equation as
Kb1 = 1.0×10^-6 = [NicH+][OH-] / [Nic]
1.0×10^-6 = x^2 / 1.85×10^-3 - x
Approximating that x is negligible compared to 1.85×10^-3 simplifies the equation to
1.0×10^-6 = x^2 / 1.85×10^-3
x = 0.0000430
x = [OH-] = 4.30×10^-5 M
From the Kb2 expression
Kb2 = 1.3×10-11 = [NicH2 2+][OH-] / [NicH+]
1.1×10^-10 = x^2 / 4.30×10^-5 - x
Approximating that x is negligible compared to 4.30×10^-5 simplifies the equation to
1.1×10^-10 = x^2 / 4.30×10^-5
x = [OH-] = 6.88×10^-8
The concentration [OH-] can be computed as
[OH-] = 4.30×10^-5 M + 6.88×10^-8 M = 4.30×10^-5 M
This shows that the second equilibrium has a negligible effect on the pH.
We can now calculate for pH:
pOH = -log [OH-] = -log (4.30×10^-5 M) = 4.37
pH = 14 - pOH = 14 - 4.37 = 9.63
physical properties for a bar of soap: hardness and color
physical properties for a book: texture and size
physical properties for tooth past: cohesiveness and taste
