Answer:
See explanation
Explanation:
We can describe electrons using four sets of quantum numbers;
principal quantum number (n)
orbital quantum number (l)
magnetic quantum number (ml)
spin quantum number (ms)
Since no two electrons in an atom can have the same value for all four quantum numbers according to Pauli exclusion theory, for the orbitals given one possible value for each quantum number is shown below;
For 1s-
n = 1, l= 0, ml = 0, ms= 1/2
For 2s-
n= 2, l =0, ml=0, ms=1/2
For 1s and 2s orbitals, there is only one possible value for ml which is zero.
a) After adding 10 mL of HCl
first, we need to get moles of (CH3)3N = molarity * volume
= 0.29 m * 0.025 L
= 0.00725M moles
then, we need to get moles of HCl = molarity * volume
= 0.3625 m * 0.01L
= 0.003625 moles
so moles of (CH3)3N remaining = moles of (CH3)3N - moles of HCl
= 0.00725 - 0.003625
= 0.003625 moles
and when the total volume = 0.01 L + 0.025L = 0.035 L
∴ [(CH3)3N] = moles remaining / total volume
= 0.003625 moles / 0.035L
= 0.104 M
when we have Pkb so we can get Kb :
pKb = - ㏒Kb
4.19 = -㏒Kb
∴Kb = 6.5 x 10^-5
when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]
and by using ICE table we assume we have:
[(CH3)3NH+] = X & [OH] = X
and [(CH3)3N] = 0.104 -X
by substitution:
∴ 6.5 x 10^-5 = X^2 / (0.104-X) by solving for X
∴X = 0.00257 M
∴[OH-] = X = 0.00257 M
∴POH = -㏒[OH]
= -㏒0.00257
= 2.5
∴ PH = 14 - POH
= 14 - 2.5
= 11.5
b) after adding 20ML of HCL:
moles of HCl = molarity * volume
= 0.3625 m * 0.02 L
= 0.00725 moles
the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:
and when the total volume = 0.02L + 0.025 = 0.045L
∴ [ (CH3)3NH+] = moles / total volume
= 0.003625 / 0.045L
= 0.08 M
when Ka = Kw / Kb
and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14
so, by substitution:
Ka = (1 x 10^-14) / (6.5 x 10^-5)
= 1.5 x 10^-10
when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]
by substitution:
∴ 1.5 x 10^-10 = X^2 / (0.08 - X) by solving for X
∴X = 3.5 x 10^-6 M
∴ [H+]= X = 3.5 x 10^-6 M
∴PH = -㏒[H+]
= -㏒(3.5 x 10^-6)
= 5.5
C) after adding 30ML of HCl:
moles of HCl = molarity * volume
= 0.3625m * 0.03L
= 0.011 moles
and when moles of (CH3)3N neutralized = 0.003625 moles
∴ moles of HCl remaining = moles HCl - moles (CH3)3N neutralized
= 0.011moles - 0.003625moles
= 0.007375 moles
when total volume = 0.025L + 0.03L
= 0.055L
∴[H+] = moles / total volume
= 0.007375 mol / 0.055L
= 0.134 M
∴PH = -㏒[H+]
= -㏒ 0.134
= 0.87
Use Avogadro's number to calculate the number of moles of Mg<span>Cl2 . The number of moles of Cl is twice as </span>much, because the ratio of Cl in MgCl2 to MgCl2 is 2:1. 3.61⋅10246.022⋅10<span>23=5.99. Therefore, there are 12 moles of Cl.</span>
