<span>it is called precipitant</span>
<h3><u>Answer;</u></h3>
pH = 12.33
<h3><u>Explanation;</u></h3>
The equation of reaction is :
LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)
Reactants left after the titrant is added;
Total Moles LiOH;
= 0.035L LiOH × (0.2moles/L)
= 0.007moles of LiOH
Moles of HCl;
= 0.023L HCl × (0.25moles/L)
= 0.00575moles HCl is the limiting reagent
Reacting amount of moles of LiOH;
= 0.0575 moles HCl *(1mole LiOH/1moles HCl)
=0.00575 moles LiOH (reacted)
Moles of LiOH left;
= 0.007moles total - 0.00575moles that react
= .00125 moles of LiOH (left)
LiOH is a strong base, which means that it ionizes completely.
0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH
LiOH(aq) --> Li+(aq) + OH-(aq)
[LiOH] = [OH-] = 0.02155 M
pOH = -log[OH-]
pOH = -log(0.02155)
pOH= 1.67
pH = 14 - pOH
pH = 14 - 1.67
pH = 12.33
Answer:
A and D are true , while B and F statements are false.
Explanation:
A) True. Since the standard gibbs free energy is
ΔG = ΔG⁰ + RT*ln Q
where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R
when the system reaches equilibrium ΔG=0 and Q=Keq
0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)
therefore the first equation also can be expressed as
ΔG = RT*ln (Q/Keq)
thus the standard gibbs free energy can be determined using Keq
B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions
C) False. From the equation presented
ΔG⁰ = (-RT*ln Keq)
ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1
for example, for a reversible reaction ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)
D) True. Standard conditions refer to
T= 298 K
pH= 7
P= 1 atm
C= 1 M for all reactants
Water = 55.6 M
